50+(2x)(3x-10)=90

Solution for 50+(2x)(3x-10)=90 equation:

50+(2x)(3x-10)=90

We move all terms to the left:

50+(2x)(3x-10)-(90)=0

We add all the numbers together, and all the variables

2x(3x-10)-40=0

We multiply parentheses

6x^2-20x-40=0

a = 6; b = -20; c = -40;
Δ = b2-4ac
Δ = -202-4·6·(-40)
Δ = 1360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{85}}{2*6}=\frac{20-4\sqrt{85}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{85}}{2*6}=\frac{20+4\sqrt{85}}{12}
$