# How do you solve 2x^2-7x+3=0?

Apr 8, 2018

$x = \frac{1}{2} \mathmr{and} x = 3$

#### Explanation:

$2 {x}^{2} - 7 x + 3 = 0$

$2 {x}^{2} - 6 x - x + 3 = 0$

$2 x \left(x - 3\right) - \left(x - 3\right) = 0$

$\left(2 x - 1\right) \left(x - 3\right) = 0$

Apr 8, 2018

The answer is $x = \frac{1}{2} , 3$, or if you prefer you can write it as $x = \frac{1}{2} \mathmr{and} x = 3$.

#### Explanation:

This problem is a quadratic equation. So you could use the tedious quadratic formula,

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

The faster method is to just factor. This problem can be easily factored but it takes some practice to be able to figure out which quadratics can and can't be factored.

So with factoring, you will get

$\left(2 x - 1\right) \left(x - 3\right) = 0$

Using the zero product property,

$2 x - 1 = 0$
$x = \frac{1}{2}$

$x - 3 = 0$
$x = 3$

So your two final answers will be $x = \frac{1}{2} , 3$ or you could write $x = \frac{1}{2} \mathmr{and} x = 3$.