Quadratic Formula

We usually try to solve quadratics by factoring. But, some quadratics can't  be factored. Do you remember how to solve those?  The quadratic formula!  Have it memorized!

On a standardized test, you'll know you need to use the quadratic formula to find the solutions to a quadratic equation when you see that the multiple choice answers include a $\pm$ and a square root! If you see that, don't even try to factor; just use the quadratic formula. 

To remind you, the quadratic formula is:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

To use the quadratic formula, you need to get your quadratic equation into standard form, so that you can pull out the $a$, $b$, and $c$ terms in order to plug them into the formula. 

Example:

What are the solutions to $3x^2-5x+2$?

This equation is already in standard form, so: $a=3$, $b=-5$, and $c=2$.  Plug these values into the quadratic formula and solve. 

$\eqalign{x=&\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\=&\dfrac{5\pm \sqrt{(-5)^2-4(3)(2)}}{2(3)}\\=&\dfrac{5 \pm \sqrt{25-24}}{6}\\=&\dfrac{5 \pm 1}{6}\\=&\dfrac{6}{6} \text{ and }\dfrac{4}{6}\\=&1 \text{ and } \dfrac{2}{3}}$

For more complicated problems, you have to reorganize the problem into standard form:

Example:

What are the solutions to $2x=20x^2-7$?

Reorganize the equation into standard form ($ax^2+bx+c$).

$\eqalign{20x&=2x^2-7\\-20x&=\quad -20x\\0&=2x^2-20x-7}$

Now: $a=2$, $b=-20$, and $c=-7$.  Plug these values into the quadratic formula and solve. 

$\eqalign{x=&\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\=&\dfrac{20\pm \sqrt{(-20)^2-4(2)(7)}}{2(2)}\\=&\dfrac{20 \pm \sqrt{400-56}}{4}\\=&\dfrac{20 \pm 344}{4}\\=&\dfrac{364}{4} \text{ and }\dfrac{-324}{4}\\=&91 \text{ and }-81}$

Some quadratics can only be solved with the quadratic formula, but the quadratic formula can be used to solve any quadratic equation, so don't hesitate to use it if you need it.