2x^2+18x-4=-11x^2-2x

Solution for 2x^2+18x-4=-11x^2-2x equation:

2x^2+18x-4=-11x^2-2x

We move all terms to the left:

2x^2+18x-4-(-11x^2-2x)=0

We get rid of parentheses

2x^2+11x^2+2x+18x-4=0

We add all the numbers together, and all the variables

13x^2+20x-4=0

a = 13; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·13·(-4)
Δ = 608
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{38}}{2*13}=\frac{-20-4\sqrt{38}}{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{38}}{2*13}=\frac{-20+4\sqrt{38}}{26}
$