With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to \(\displaystyle 0\):

Thus, simplify \(\displaystyle x^2 + 4x + 24 = 6 - 5x\) into:

\(\displaystyle x^2 + 9x + 18 = 0\)

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can. Factor the quadratic expression:

\(\displaystyle (x+3)(x+6)=0\)

Now, remember that you merely need to set each group equal to \(\displaystyle 0\). This gives you the two values for \(\displaystyle x\):

\(\displaystyle x+3=0\); therefore \(\displaystyle x=-3\)

Likewise, for the other group, \(\displaystyle x=-6\)

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to \(\displaystyle 0\):

Thus, simplify \(\displaystyle 2x^2 + 19x + 10 = 4 - x^2\) into:

\(\displaystyle 3x^2 + 19x + 6 = 0\)

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it can, though we are sometimes a bit intimidated by \(\displaystyle x^2\) terms that have a coefficient like this. Factor the quadratic expression:

\(\displaystyle (3x+1)(x+6)=0\)

If you FOIL this out, you will see your original equation.

Now, remember that you merely need to set each group equal to \(\displaystyle 0\). This gives you the two values for \(\displaystyle x\):

\(\displaystyle 3x+1=0\)

\(\displaystyle 3x=-1\)

\(\displaystyle x=\frac{-1}{3}\)

For the other group, you get \(\displaystyle x=-6\).

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to \(\displaystyle 0\):

Thus, simplify \(\displaystyle x^2 + 16x - 5 = 10 - 4x\) into:

\(\displaystyle x^2 + 20x - 15 = 0\)

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it cannot be easily factored. Therefore, you should use the quadratic formula. Recall that its general form is:

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

For our data, \(\displaystyle a=1\), \(\displaystyle b=20\), and \(\displaystyle c=-15\).

Thus, we have:

\(\displaystyle \frac{-20\pm\sqrt{20^2-(4*1*-15)}}{2*1}\)

Simplifying, this is:

\(\displaystyle \frac{-20\pm\sqrt{400-(-60)}}{2}\)

\(\displaystyle \frac{-20\pm\sqrt{400+60}}{2}\)

\(\displaystyle \frac{-20\pm\sqrt{460}}{2}\)

Now, simplify.

\(\displaystyle \frac{-20\pm\sqrt{2*2*5*23}}{2}=\frac{-20\pm2\sqrt{115}}{2}=-10\pm\sqrt{115}\)

With quadratic equations, you should always start by getting all of your terms to one side of the equation, setting this equal to \(\displaystyle 0\):

Thus, simplify \(\displaystyle 7x^2 + 3x +25 = 4x - 10\) into:

\(\displaystyle 7x^2 -x +35 = 0\)

Now, the next question you need to ask yourself is, "Can this be factored?" In this case, it cannot be easily factored. Therefore, you should use the quadratic formula. Recall that its general form is:

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

For our data, \(\displaystyle a=7\), \(\displaystyle b=-1\), and \(\displaystyle c=35\).

Thus, we have:

\(\displaystyle \frac{-(-1)\pm\sqrt{(-1)^2-(4*7*35)}}{2*-1}\)

Simplifying, this is:

\(\displaystyle \frac{1\pm\sqrt{1-(980)}}{-2}\)

Since \(\displaystyle 1-980\) is negative, you know that there is no real solution (given the problems arising from the negative square root)!

To solve, we must set it equal to zero. The above expression is of the form \(\displaystyle ax^{2}+bx+c\), so we can use the quadratic formula:

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

to solve for the roots which are \(\displaystyle 2\) and \(\displaystyle 3\)

We can check by plugging the roots into the expression and making sure that it equals zero.