x^2+2x-48=3x+8

Solution for x^2+2x-48=3x+8 equation:

x^2+2x-48=3x+8

We move all terms to the left:

x^2+2x-48-(3x+8)=0

We get rid of parentheses

x^2+2x-3x-8-48=0

We add all the numbers together, and all the variables

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