# How do you find the zeros, if any, of y= 2x^2 -12x+ 23 using the quadratic formula?

Mar 9, 2016

$\Delta < 0$
$\therefore \cancel{\exists} x \in \mathbb{R}$

The equation has not solutions in $\mathbb{R}$

or, if the excercise allow it

${x}_{1 , 2} = 3 \pm i \frac{\sqrt{10}}{2} \in \mathbb{C}$

#### Explanation:

Given

$y = a {x}^{2} + b x + c = 2 {x}^{2} - 12 x + 23 = 0$

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

where:

$a = 2$
$b = - 12$
$c = 23$

$\therefore {x}_{1 , 2} = \frac{12 \pm \sqrt{144 - 4 \cdot 2 \cdot 23}}{4} = \frac{12 \pm \sqrt{144 - 4 \cdot 2 \cdot 23}}{4} = \frac{12 \pm \sqrt{144 - 184}}{4} = \frac{12 \pm \sqrt{- 40}}{4}$

Now, $\Delta < 0$ therefore $\cancel{\exists} x \in \mathbb{R}$

You have solutions in $\mathbb{C}$, if the exercise allows them

${x}_{1 , 2} = \frac{12 \pm \sqrt{- 1} \sqrt{40}}{4} = \frac{12 \pm i \sqrt{4 \cdot 10}}{4} =$
$= \frac{{\cancel{12}}^{6} \pm i \cancel{2} \sqrt{10}}{\cancel{4}} ^ 2 = 3 \pm i \frac{\sqrt{10}}{2}$