# How do you solve (2x ^ { 2} + 3x - 5) + ( x ^ { 2} + 11x - 1) = 180?

Jul 18, 2018

#### Answer:

${x}_{1 , 2} = - \frac{7}{3} \pm \frac{\sqrt{607}}{3}$

#### Explanation:

Combining like Terms

$3 {x}^{2} + 14 x - 6 = 180$

adding $- 180$

$3 {x}^{2} + 14 x - 186 = 0$

dividing by $3$

${x}^{2} + \frac{14}{3} x - 62 = 0$

so we get

by the quadratic formula
${x}^{2} + p x + q = 0$

${x}_{1 , 2} = - \frac{p}{2} \pm \sqrt{{p}^{2} / 4 - q}$

with $p = \frac{14}{3} , q = - 162$

${x}_{1 , 2} = - \frac{7}{3} \pm \setminus \frac{\sqrt{607}}{3}$

Jul 18, 2018

Collect like terms

$2 {x}^{2} + 3 x - 5 + {x}^{2} + 11 x - 1 = 180$

$3 {x}^{2} + 14 x - 6 - 180 = 0$

$3 {x}^{2} + 14 x - 186 = 0$

Use the formula

$x = \frac{- 14 \setminus \pm \sqrt{{14}^{2} + 4 \times 3 \times 186}}{2 \times 3}$

$x = \frac{- 14 \setminus \pm \sqrt{2428}}{6}$

$x = \frac{- 7 + \sqrt{607}}{3}$ or $x = \frac{- 7 - \sqrt{607}}{3}$

$x = 5.87912333 \mathmr{and} x = - 10.54579$