# How do you solve 2x^2 - 3x + 4 = 0 using the quadratic formula?

Jan 20, 2017

$x = 3 \pm \frac{\sqrt{7}}{2 \sqrt{2}}$

#### Explanation:

$f \left(x\right) = {\left(x \sqrt{2} - \frac{3}{\sqrt{2}}\right)}^{2} + 4 - \frac{9}{4} = 0$
if $x \sqrt{2} - \frac{3}{\sqrt{2}} = \pm \frac{\sqrt{7}}{4}$
or $x = 3 \pm \frac{\sqrt{7}}{2 \sqrt{2}}$

Jan 20, 2017

$x = \frac{3}{4} \pm \frac{\sqrt{23}}{4} i$

#### Explanation:

Given:

$2 {x}^{2} - 3 x + 4 = 0$

Note that this is of the form:

$a {x}^{2} + b x + c = 0$

with $a = 2$, $b = - 3$ and $c = 4$

The discriminant $\Delta$ of this quadratic is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - 4 \left(2\right) \left(4\right) = 9 - 32 = - 23$

Since $\Delta < 0$ we can tell that this quadratic has no Real zeros - only Complex ones.

We can still use the quadratic formula to find them:

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

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{3 \pm \sqrt{- 23}}{4}$

$\textcolor{w h i t e}{x} = \frac{3 \pm \sqrt{23} i}{4}$

$\textcolor{w h i t e}{x} = \frac{3}{4} \pm \frac{\sqrt{23}}{4} i$

Jan 20, 2017

$2 {x}^{2} - 3 x + 4 = 0$

The solutions to $a {x}^{2} + b x + c = 0$ are given by

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

In this case, we get

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

$= \frac{3 \pm \sqrt{9 - 32}}{4}$

$= \frac{3 \pm \sqrt{- 23}}{4}$

The solutions are imaginary

$\frac{3}{4} \pm \frac{\sqrt{23}}{4} i$