# How do you solve using the completing the square method 10x^2 = 4x + 7?

Jul 9, 2017

${x}_{1} = \frac{2 + \sqrt{74}}{10} \mathmr{and} {x}_{2} = \frac{2 - \sqrt{74}}{10}$

#### Explanation:

$10 {x}^{2} = 4 x + 7$

$10 {x}^{2} - 4 x - 7 = 0$

$100 {x}^{2} - 40 x - 70 = 0$

$100 {x}^{2} - 40 x + 4 - 74 = 0$

${\left(10 x - 2\right)}^{2} - {\left(\sqrt{74}\right)}^{2} = 0$

${\left(10 x - 2\right)}^{2} = {\left(\sqrt{74}\right)}^{2}$

Hence ${x}_{1} = \frac{2 + \sqrt{74}}{10}$ and ${x}_{2} = \frac{2 - \sqrt{74}}{10}$

Jul 9, 2017

$x = \frac{1}{5} \pm \sqrt{\frac{37}{50}}$

#### Explanation:

$10 {x}^{2} - 4 x = 7$
Divide both sides by 10:
${x}^{2} - \frac{4 x}{10} = \frac{7}{10}$
${x}^{2} - \left(2 \frac{x}{5}\right) = \frac{7}{10}$
$\left({x}^{2} - \frac{2 x}{5}\right) + \frac{1}{25} = \frac{7}{10} + \frac{1}{25}$
${\left(x - \frac{1}{5}\right)}^{2} = \frac{37}{50}$
$\left(x - \frac{1}{5}\right) = \pm \sqrt{\frac{37}{50}}$
$x = \frac{1}{5} \pm \sqrt{\frac{37}{50}}$