# How do you solve the quadratic using the quadratic formula given x^2-3x+10=0?

Nov 9, 2016

$0 = {x}^{2} - 3 x + 10$

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

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(10\right)}}{2 \left(1\right)}$

Simplify:
$x = \frac{3 \pm \sqrt{- 31}}{2}$

$x = \frac{3 \pm i \sqrt{31}}{2}$

Nov 9, 2016

$x = \frac{3 + \sqrt{31} i}{2} , \frac{3 - \sqrt{31} i}{2}$

#### Explanation:

${x}^{2} - 3 x + 10 = 0$ is a quadratic equation in the form $a {x}^{2} + b x + c$, where $a = 1$, $b = - 3$, and $c = 10$.

$x = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$
$x = \frac{3 \pm \sqrt{9 - 40}}{2}$
$x = \frac{3 \pm \sqrt{- 31}}{2}$
$x = \frac{3 + \sqrt{31} i}{2} , \frac{3 - \sqrt{31} i}{2}$