# How do you solve using completing the square method 3x^2 + 5x = -x + 4?

Jun 26, 2018

$x = - 1 \pm \sqrt{\frac{7}{3}}$

#### Explanation:

$3 {x}^{2} + 5 x = - x + 4$

$3 {x}^{2} + 6 x = 4$

${\left(\sqrt{3} x\right)}^{2} + 2 \cdot \sqrt{3} \cdot \sqrt{3} x + {\left(\sqrt{3}\right)}^{2} = 4 + {\left(\sqrt{3}\right)}^{2}$

${\left(\sqrt{3} x + \sqrt{3}\right)}^{2} = 7$

$\left(\sqrt{3} x + \sqrt{3}\right) = \pm \left(\sqrt{7}\right)$

$x + 1 = \sqrt{\frac{7}{3}}$

$x = - 1 \pm \sqrt{\frac{7}{3}}$

Jun 26, 2018

$x = - 1 \pm \sqrt{\frac{7}{3}}$

#### Explanation:

$3 {x}^{2} + 5 x = - x + 4$

$3 {x}^{2} + 6 x = 4$

${x}^{2} + 2 x = \frac{4}{3}$

${x}^{2} + 2 x + 1 = \frac{4}{3} = 1 = \frac{7}{3}$

${\left(x + 1\right)}^{2} = {\left(\sqrt{\frac{7}{3}}\right)}^{2}$

$x = 1 = \pm \sqrt{\frac{7}{3}}$

$x = - 1 \pm \sqrt{\frac{7}{3}}$