We will call the constant  that goes in the blank \(\displaystyle N\). The equation becomes

\(\displaystyle 3x^{2} + 7x = N\)

Write the quadratic equation in standard form \(\displaystyle ax^{2} + bx+ c = 0\) by subtracting \(\displaystyle N\) from both sides:

\(\displaystyle 3x^{2} + 7x - N = N - N\)

\(\displaystyle 3x^{2} + 7x - N = 0\)

The solution set comprises exactly one rational solution if and only if the discriminant \(\displaystyle b^{2} - 4ac\) is equal to 0. Setting \(\displaystyle a = 3, b= 7 , c= -N\). and substituting in the equation:

\(\displaystyle b^{2} - 4ac = 0\)

\(\displaystyle 7^{2} - 4(3)(-N) = 0\)

Solving for \(\displaystyle N\):

\(\displaystyle 49 - (-12N) = 0\)

\(\displaystyle 49 + 12 N = 0\)

\(\displaystyle 49 + 12 N- 49 = 0 - 49\)

\(\displaystyle 12N = -49\)

\(\displaystyle 12N\div 12 = -49 \div 12\)

\(\displaystyle N = - \frac{49}{12}\),

the correct response.

We will call the constant  that goes in the blank \(\displaystyle N\). The equation becomes 

\(\displaystyle N x^{2} + 16 = 7x\)

Write the quadratic equation in standard form \(\displaystyle ax^{2} + bx+ c = 0\) by subtracting \(\displaystyle 7x\) from both sides:

\(\displaystyle N x^{2} + 16 - 7x = 7x - 7x\)

\(\displaystyle N x^{2}- 7x + 16 = 0\)

The solution set comprises exactly one rational solution if and only if the discriminant \(\displaystyle b^{2} - 4ac\) is equal to 0. Setting \(\displaystyle a =N, b= -7 , c= 16\). and substituting in the equation:

\(\displaystyle b^{2} - 4ac = 0\)

\(\displaystyle (-7)^{2} - 4 (N)(16)= 0\)

\(\displaystyle 49 - 64N= 0\)

Solving for \(\displaystyle N\):

\(\displaystyle 49 - 64N+ 64N = 0 + 64N\)

\(\displaystyle 49 = 64N\)

\(\displaystyle 49 \div 64 = 64N \div 64\)

\(\displaystyle N = \frac{49}{64}\),

the correct response.

We will call the constant  that goes in the blank \(\displaystyle N\). The equation becomes 

\(\displaystyle 3 x^{2} + 48 = Nx\)

Write the quadratic equation in standard form \(\displaystyle ax^{2} + bx+ c = 0\) by subtracting \(\displaystyle Nx\) from both sides:

\(\displaystyle 3 x^{2} + 48 - Nx = Nx - Nx\)

\(\displaystyle 3 x^{2} - Nx + 48 = 0\)

The solution set comprises exactly one rational solution if and only if the discriminant \(\displaystyle b^{2} - 4ac\) is equal to 0. Setting \(\displaystyle a = 3, b = -N , c= 48\). and substituting in the equation:

\(\displaystyle b^{2} - 4ac = 0\)

\(\displaystyle (-N)^{2} - 4 (3)(48)= 0\)

\(\displaystyle N^{2} - 576= 0\)

Solving for \(\displaystyle N\):

\(\displaystyle N^{2} - 576+ 576 = 0+ 576\)

\(\displaystyle N^{2} = 576\)

\(\displaystyle N = \pm \sqrt{576}\)

\(\displaystyle N = \pm 24\),

that is, either \(\displaystyle N = 24\) or \(\displaystyle N = -24\).

\(\displaystyle -24\) is not a choice, but 24 is; this is the correct response.

Multiply both sides of the top equation by 7:

\(\displaystyle 0.7 x + 0.4 y = 0.07\)

\(\displaystyle 7(0.7 x + 0.4 y )= 7(0.07)\)

\(\displaystyle 7(0.7 x )+7( 0.4 y )= 7(0.07)\)

\(\displaystyle 4.9 x+2.8 y = 0.49\)

Multiply both sides of the bottom equation by \(\displaystyle -4\):

\(\displaystyle 0.4 x+ 0.7y = -0.95\)

\(\displaystyle -4 (0.4 x+ 0.7y )=-4 ( -0.95)\)

\(\displaystyle -4(0.4 x)+ (-4)(0.7y )=-4( -0.95)\)

\(\displaystyle -1.6x-2.8y = 3.8\)

Add both sides of the equations to eliminate the \(\displaystyle y\) terms:

\(\displaystyle 4.9 x+2.8 y = 0.49\)

\(\displaystyle \underline{-1.6x-2.8y = 3.8}\)

\(\displaystyle 3.3x\)                 \(\displaystyle = 4.29\)

Solve for \(\displaystyle x\):

\(\displaystyle 3.3x \div 3.3 = 4.29 \div 3.3\)

\(\displaystyle x = 1.3\)

This is a quadratic equation because the leading term is of degree \(\displaystyle 2\); hence, its solutions can be easily calculated via the quadratic formula:

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

In order to use the quadratic formula to solve a quadratic equation, you must identify the values of the coefficients \(\displaystyle a\), \(\displaystyle b\), and \(\displaystyle c\), substitute them into the quadratic formula, and perform the arithmetical calculations to yield one, two, or no real number solutions for \(\displaystyle x\).

In this case, \(\displaystyle a=2\), \(\displaystyle b=-6\), and \(\displaystyle c=-36\). Hence, the quadratic formula yields

\(\displaystyle x=\frac{-(-6)\pm\sqrt{(-6)^2-4(2)(-36)}}{2(2)}\)

\(\displaystyle =\frac{6\pm\sqrt{36+288}}{4}\)

\(\displaystyle =\frac{6\pm\sqrt{324}}{4}\)

\(\displaystyle =\frac{6\pm18}{4}\)

\(\displaystyle x=6, x=-3\)

Hence, this equation has two real solutions: \(\displaystyle -3\) and \(\displaystyle 6\).

With quadratic equations, always begin by getting it into standard form:

\(\displaystyle Ax^2+Bx+C=0\)

Therefore, take our equation:

\(\displaystyle 3x^2-5x=2x+6\)

And rewrite it as:

\(\displaystyle 3x^2-7x-6=0\)

You could use the quadratic formula to solve this problem.  However, it is possible to factor this if you are careful.  Factored, the equation can be rewritten as:

\(\displaystyle (3x+2)(x-3)=0\)

Now, either one of the groups on the left could be \(\displaystyle 0\) and the whole equation would be \(\displaystyle 0\).  Therefore, you set up each as a separate equation and solve for \(\displaystyle x\):

\(\displaystyle 3x+2=0\)

\(\displaystyle 3x=-2\)

\(\displaystyle x=\frac{-2}{3}\)

OR

\(\displaystyle x-3=0\)

\(\displaystyle x=3\)

The sum of these values is:

\(\displaystyle 3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}\)

Tommy throws a rock off a 10 meter ledge at a speed of 3 meters/second. To calculate when the rock hits the ground first identify what is known.

Using the equation 

\(\displaystyle s(t)=-gt^2+v_0(t)+h_0\)

where

\(\displaystyle \\g=4.9 \\v_0=\text{speed} \\h_0=\text{starting height}\)

it is known that,

\(\displaystyle v_0=3\frac{ \text{ m}}{\text{ s}}\)

\(\displaystyle h_0=10 \text{ m}\)

Substituting the given values into the position \(\displaystyle s(t)\) equation looks as follows.

\(\displaystyle s(t)=-4.9t^2+3(t)+10\)

Now to calculate when the rock hits the ground, find the \(\displaystyle t\) value that results in \(\displaystyle s(t)=0\).

\(\displaystyle 0=-4.9t^2+3t+10\)

Use graphing technology to graph \(\displaystyle s(t)\).

It appears that the rock hits the ground approximately 1.75 seconds after Tommy throws it.

The first step is to set it equal to zero.

\(\displaystyle 0=x^2+qx+r\)

Now we will use the quadratic formula.

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\)

In this case \(\displaystyle a=1\), \(\displaystyle b=q\), \(\displaystyle c=r\)

\(\displaystyle x=\frac{-q\pm\sqrt{q^2-4\cdot 1\cdot r}}{2\cdot 1}\)

\(\displaystyle x=\frac{-q\pm\sqrt{q^2-4 r}}{2}\)