(2x-2)(x+3)=42

Solution for (2x-2)(x+3)=42 equation:

(2x-2)(x+3)=42

We move all terms to the left:

(2x-2)(x+3)-(42)=0

We multiply parentheses ..

(+2x^2+6x-2x-6)-42=0

We get rid of parentheses

2x^2+6x-2x-6-42=0

We add all the numbers together, and all the variables

2x^2+4x-48=0

a = 2; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·2·(-48)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*2}=\frac{16}{4}
=4
$