x^2+18x=4x-40

Solution for x^2+18x=4x-40 equation:

x^2+18x=4x-40

We move all terms to the left:

x^2+18x-(4x-40)=0

We get rid of parentheses

x^2+18x-4x+40=0

We add all the numbers together, and all the variables

x^2+14x+40=0

a = 1; b = 14; c = +40;
Δ = b2-4ac
Δ = 142-4·1·40
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*1}=\frac{-8}{2}
=-4
$