# How do you use the quadratic formula to solve 2x^2-3x+2=0?

Mar 30, 2018

No real solutions. Imaginary solutions: $x = \frac{3 \pm \sqrt{7} i}{4}$

#### Explanation:

The quadratic formula allows us to solve a quadratic equation in the form $a {x}^{2} + b x + c$ for $x$ using the following:

$x = \frac{- b \pm \sqrt{{b}^{4} - 4 a c}}{2 a}$. So, generally, there will be two solutions.

For $2 {x}^{2} - 3 x + 2 , a = 2 , b = - 3 , c = 2$, and so

$x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - \left(4\right) \left(2\right) \left(2\right)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 16}}{4} = \frac{3 \pm \sqrt{- 7}}{4}$

Due to the negative root, there will be no real solutions, but the imaginary solutions would be

$x = \frac{3 \pm \sqrt{7} i}{4}$