x^2+4x-40=3x+16

Solution for x^2+4x-40=3x+16 equation:

x^2+4x-40=3x+16

We move all terms to the left:

x^2+4x-40-(3x+16)=0

We get rid of parentheses

x^2+4x-3x-16-40=0

We add all the numbers together, and all the variables

x^2+x-56=0

a = 1; b = 1; c = -56;
Δ = b2-4ac
Δ = 12-4·1·(-56)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*1}=\frac{14}{2}
=7
$