2x^2+2x-40x-4=180

Solution for 2x^2+2x-40x-4=180 equation:

2x^2+2x-40x-4=180

We move all terms to the left:

2x^2+2x-40x-4-(180)=0

We add all the numbers together, and all the variables

2x^2-38x-184=0

a = 2; b = -38; c = -184;
Δ = b2-4ac
Δ = -382-4·2·(-184)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-54}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+54}{2*2}=\frac{92}{4}
=23
$