# How do you find the roots, real and imaginary, of y=x^2 - 3x + 2  using the quadratic formula?

Dec 9, 2015

You can find all the roots of $y = {x}^{2} - 3 x + 2$ by "plugging in" values from the equation into the quadratic formula to find that the roots are $\left(2 , 0\right)$ and $\left(1 , 0\right)$.

#### Explanation:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The equation representing any given quadratic, or parabola, can be:

$y = a {x}^{2} + b x + c$

So all we need to do is assign values of our equation to their respective variables, and then substitute them into the quadratic formula to solve. In the equation $y = {x}^{2} - 3 x + 2$:

$a = 1$, $b = \left(- 3\right)$, and $c = 2$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ Given Quadratic Formula

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(2\right)}}{2 \left(1\right)}$ Substitute

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

$x = \frac{3 \pm \sqrt{9 - 8}}{2}$ Simplify Some More

$x = \frac{3 \pm \sqrt{1}}{2}$ Simplify A Little More

$x = \frac{3 \pm 1}{2}$ Simplify Even More

$x = \frac{3 + 1}{2}$, OR $x = \frac{3 - 1}{2}$ Isolate Both Possible Roots

$x = 2$ OR $x = 1$ Simplify (For the Last Time..)

This means that the roots, or x-intercepts, of our quadratic equation are at the points: $\left(2 , 0\right)$ and $\left(1 , 0\right)$.