# How do you solve 2x^2 - 3x + 1 = 0?

Apr 4, 2018

$x = \frac{1}{2} , 1$

Here's how I solved it:

#### Explanation:

You solve this by using the quadratic formula. This equation is written in standard form, or $a {x}^{2} + b x + c = 0$. The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

So let's plug those numbers in and solve for $x$:
$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(2\right) \left(1\right)}}{2 \left(2\right)}$

$x = \frac{3 \pm \sqrt{9 - 8}}{4}$

$x = \frac{3 \pm \sqrt{1}}{4}$

$x = \frac{3 \pm 1}{4}$

So $x = \frac{3 + 1}{4}$ and $\frac{3 - 1}{4}$.

$x = \frac{4}{4} \to x = 1$

$x = \frac{2}{4} \to x = \frac{1}{2}$

$x = \frac{1}{2} , 1$

Hope this helps!