X^2+4x=2x+18

Solution for X^2+4x=2x+18 equation:

X^2+4X=2X+18

We move all terms to the left:

X^2+4X-(2X+18)=0

We get rid of parentheses

X^2+4X-2X-18=0

We add all the numbers together, and all the variables

X^2+2X-18=0

a = 1; b = 2; c = -18;
Δ = b2-4ac
Δ = 22-4·1·(-18)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{19}}{2*1}=\frac{-2-2\sqrt{19}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{19}}{2*1}=\frac{-2+2\sqrt{19}}{2}
$