# How do you solve this quadratic equation 2x^2- 10+ 7=0?

Jun 3, 2018

$x = \pm \frac{1}{2} \sqrt{6}$

#### Explanation:

$2 {x}^{2} - 10 + 7 = 0$
Add 3 on both sides to move the constants to the right side:
$2 {x}^{2} - 10 + 7 + 3 = 0 + 3$
$2 {x}^{2} = 3$
Divide both sides with 2:
${x}^{2} = \frac{3}{2}$
$x = \pm \sqrt{\frac{3}{2}} = \pm \frac{1}{2} \sqrt{3} \sqrt{2} = \pm \frac{1}{2} \sqrt{6}$
(if we multiply the right side with $\frac{\sqrt{2}}{\sqrt{2}}$)

Jun 3, 2018

$x = \frac{5 \pm \sqrt{11}}{2}$

#### Explanation:

if $2 {x}^{2} - 10 x + 7 = 0$

You can use the quadratic formula:

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

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

In this case $a = 2$ , $b = - 10$ , $c = 7$

$\implies x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - \left(4 \cdot 2 \cdot 7\right)}}{2 \cdot 2}$

$\implies x = \frac{10 \pm \sqrt{44}}{4}$

$\implies x = \frac{10 \pm 2 \sqrt{11}}{4}$

color(red)(=> x = (5 pm sqrt(11))/2