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

Question

2107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-18T08:45:41

$x_{1, 2} = - \frac{7}{3} \pm \frac{\sqrt{607}}{3}$ $x_(1,2)=-7/3pmsqrt(607)/3$

Combining like Terms

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

adding $- 180$ $-180$

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

dividing by $3$ $3$

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

so we get

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

$x_{1, 2} = - \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - q}$ $x_(1,2)=-p/2pmsqrt(p^2/4-q)$

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

$x_{1, 2} = - \frac{7}{3} \pm \frac{\sqrt{607}}{3}$ $x_(1,2)=-7/3pm\sqrt(607)/3$

Answer 2 · 2018-07-18T08:59:47

Collect like terms

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

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

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

Use the formula

$x = \frac{- 14 \pm \sqrt{14^{2} + 4 \times 3 \times 186}}{2 \times 3}$ $x=(-14\pmsqrt(14^2+4xx3xx186))/(2xx3)$

$x = \frac{- 14 \pm \sqrt{2428}}{6}$ $x=[-14\pmsqrt(2428)]/6$

$x = \frac{- 7 + \sqrt{607}}{3}$ $x=[-7+sqrt607]/3$ or $x = \frac{- 7 - \sqrt{607}}{3}$ $x=[-7-sqrt607]/3$

$x = 5.87912333 or x = - 10.54579$ $x=5.87912333 or x=-10.54579$

How do you solve (2x ^ { 2} + 3x - 5) + ( x ^ { 2} + 11x - 1) = 180(2x2+3x−5)+(x2+11x−1)=180(2x ^ { 2} + 3x - 5) + ( x ^ { 2} + 11x - 1) = 180?