\(\displaystyle 2x^{2}+15x+25=0\)

1) The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

2) Split up the middle term to make factoring by grouping possible.

\(\displaystyle 2x^{2}+10x+5x+25=0\)

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

3) Pull out the common factors from both groups, "2x" from the first and "5" from the second.

\(\displaystyle 2x(x+5)+5(x+5)=0\)

4) Factor out the "(x+5)" from both terms.

\(\displaystyle (2x+5)(x+5)=0\)

5) Set each parenthetical expression equal to zero and solve.

2x + 5 = 0,  x = –5/2

x + 5 = 0, x = –5

The quadratic formula is as follows:

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

We will start by finding the values of the coefficients of the given equation:

\(\displaystyle 2x^{2}+3x-9\)

Quadratic equations may be written in the following format:

\(\displaystyle ax^{2}+bx+c\)

In our case, the values of the coefficients are:

\(\displaystyle a=2\)

\(\displaystyle b=3\)

\(\displaystyle c=-9\)

Substitute the coefficient values into the quadratic equation:

\(\displaystyle x=\frac{-3 \pm \sqrt{3^{2}-4\left ( 2 \right )\left ( -9 \right )}}{2\left ( 2 \right )}\)

\(\displaystyle x=\frac{-3 \pm \sqrt{9-(-72)}}{4}\)

\(\displaystyle x=\frac{-3 \pm \sqrt{81}}{4}\)

After simplifying we are left with:

\(\displaystyle x= \frac{-3\pm 9}{4}\)

leaving us with our two solutions:

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

Use the quadratic equation: 

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

and use the general formula

\(\displaystyle ax^2+bx+c\)

to find the coefficients.

In our case, \(\displaystyle 2x^{2}-15x+28=0\) thus, \(\displaystyle a=2\), \(\displaystyle b=-15\), \(\displaystyle c=28\).

Plugging our values into the quadratic equation we are able to solve for \(\displaystyle x\).

\(\displaystyle x=\frac{-(-15))\pm \sqrt{(-15)^{2}-4(2)(28)}}{2(2))}\)

\(\displaystyle x=\frac{15\pm \sqrt{225-224}}{4}\)

\(\displaystyle x=\frac{15\pm 1}{4}\)

\(\displaystyle x=4,\,x=\frac{7}{2}\)

Our first step is to factor the expression. We must find two numbers which have a sum of -4 and a product of -32. -32 can be factored into +/-2 and -/+16, or +/-4 and -/+8. The only combination that satisfies the product and sum is +4 and -8. 

This means we can factor our expression to \(\displaystyle (x+4)(x-8)\). 

To find the zeroes, we simply need to find where our expression equals zero. And we can find this easily by determining where (x+4) and (x-8) each equal zero.

The zeroes of our expression, then, are at -4 and 8. 

\(\displaystyle (-4+4)(-4-8)=(0)(-12)=0\)

and

\(\displaystyle (8+4)(8-8)=(4)(0)=0\)

1) Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

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

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

\(\displaystyle x=\frac{4\pm \sqrt{16+16}}{2}\)

\(\displaystyle x=\frac{4\pm \sqrt{32}}{2}\)

\(\displaystyle x=\frac{4\pm \sqrt{16\cdot 2}}{2}=\frac{4\pm \sqrt{16}\cdot \sqrt{2}}{2}=\frac{4\pm4\sqrt{2}}{2}\)

\(\displaystyle x=2\pm2\sqrt{2}\)

\(\displaystyle x^2+4x=12\)

Put the quadratic in standard form:

\(\displaystyle x^2+4x-12=0\)

Factor:

\(\displaystyle (x+6)(x-2)\)

\(\displaystyle Solutions=-6\hspace{1mm}and\hspace{1mm}2\)

An easy way to factor (and do so with less trial and error) is to think of what two numbers could multiply to equal "c", but add to equal "b". These letters come from the designations in the standard form of a quadratic equation: \(\displaystyle ax^2+bx+c\). As you can see the product of -6 and 2 is -12 and they both add to 4.

Use the quadratic formula to solve the equation.

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

\(\displaystyle a=3\)

\(\displaystyle b=12\)

\(\displaystyle c=2\)

Plug in these values and solve.

\(\displaystyle x=\frac{-12\pm \sqrt{12^{2}-4(3)(2)}}{2(3)}\)

\(\displaystyle x=\frac{-12\pm \sqrt{144-24}}{6}\)

\(\displaystyle x=\frac{-12\pm \sqrt{120}}{6}\)

\(\displaystyle x=\frac{-6\pm \sqrt{30}}{3}\)

We start by subtracting \(\displaystyle 12\) on both sides in order to get a quadratic expression on one side of the equation:

\(\displaystyle x^{2}+4x-12=0\)

We can factor this into:

\(\displaystyle (x-2)(x+6)=0\)

Thus, either \(\displaystyle x-2=0\) or \(\displaystyle x+6=0\).

So, \(\displaystyle x=2\) or \(\displaystyle x=-6\)

To solve for x, we must first simplify the trinomial into two binomials.

To simplify the trinomial, its general form given by \(\displaystyle ax^2+bx+c\), we must find factors of \(\displaystyle ac\) that when added give us \(\displaystyle b\). For our trinomial, \(\displaystyle ac=36\) and the two factors that add together to get \(\displaystyle b=-13\) are \(\displaystyle -9\) and \(\displaystyle -4\).

Now, using the two factors, we can rewrite \(\displaystyle bx\) as the sum of the two factors each multiplied by x:

\(\displaystyle 3x^2-9x-4x+12=0\)

Next, we group the first two and last two terms together and factor each of the groups:

\(\displaystyle 3x(x-3)-4(x-3)=0\)

Now, simplify further:

\(\displaystyle (3x-4)(x-3)=0\)

Finally, set each of these binomials equal to zero and solve for x:

\(\displaystyle x=3, \frac{4}{3}\)

Write the quadratic equation.

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

The quadratic is in the form:  \(\displaystyle y=ax^2 +bx+c\)

Substitute the known coefficients.

\(\displaystyle x=\frac{-2\pm \sqrt{(2)^2-4(3)(1)}}{2(3)}\)

Simplify the numerator and denominator.

\(\displaystyle x=\frac{-2\pm \sqrt{4-12}}{6}\)

\(\displaystyle x=\frac{-2\pm \sqrt{-8}}{6}\)

\(\displaystyle x=\frac{-2\pm i\sqrt{8}}{6} = \frac{-2\pm i \cdot 2\sqrt2}{6}=\frac{-1\pm i \sqrt2}{3}\)

Because we have a negative value inside the square root, we will have complex roots instead of real roots.  The roots are imaginary.

The answer is:  \(\displaystyle \textup{No real roots.}\)