x^2+4x=2x^2-x-6

Solution for x^2+4x=2x^2-x-6 equation:

x^2+4x=2x^2-x-6

We move all terms to the left:

x^2+4x-(2x^2-x-6)=0

We get rid of parentheses

x^2-2x^2+4x+x+6=0

We add all the numbers together, and all the variables

-1x^2+5x+6=0

a = -1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-1)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-1}=\frac{2}{-2}
=-1
$