# How do you solve the equation 3x^2-5x-8=2x-3 by completing the square?

Jul 24, 2017

$x = \frac{1}{6} \left(7 \pm \sqrt{109}\right)$

#### Explanation:

$3 {x}^{2} - 5 x - 8 = 2 x - 3 \mathmr{and} 3 {x}^{2} - 5 x - 8 - 2 x + 3 = 0$ or

$3 {x}^{2} - 7 x - 5 = 0 \mathmr{and} 3 \left({x}^{2} - \frac{7}{3} x\right) - 5$ or

$3 \left({x}^{2} - \frac{7}{3} x + {\left(\frac{7}{6}\right)}^{2}\right) - 3 \cdot \frac{49}{36} - 5 = 0$

$3 {\left(x - \frac{7}{6}\right)}^{2} - \frac{49}{12} - 5 \mathmr{and} 3 {\left(x - \frac{7}{6}\right)}^{2} = \frac{109}{12}$ or

${\left(x - \frac{7}{6}\right)}^{2} = \frac{109}{3 \cdot 12} \mathmr{and} \left(x - \frac{7}{6}\right) = \pm \sqrt{\frac{109}{36}}$ or

$x = \frac{7}{6} \pm \frac{\sqrt{109}}{6} = \frac{1}{6} \left(7 \pm \sqrt{109}\right)$ [Ans]