2x^2-2x-8=3x+4

Solution for 2x^2-2x-8=3x+4 equation:

2x^2-2x-8=3x+4

We move all terms to the left:

2x^2-2x-8-(3x+4)=0

We get rid of parentheses

2x^2-2x-3x-4-8=0

We add all the numbers together, and all the variables

2x^2-5x-12=0

a = 2; b = -5; c = -12;
Δ = b2-4ac
Δ = -52-4·2·(-12)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*2}=\frac{-6}{4}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*2}=\frac{16}{4}
=4
$