(3x+100)(5x-20)=90

Solution for (3x+100)(5x-20)=90 equation:

(3x+100)(5x-20)=90

We move all terms to the left:

(3x+100)(5x-20)-(90)=0

We multiply parentheses ..

(+15x^2-60x+500x-2000)-90=0

We get rid of parentheses

15x^2-60x+500x-2000-90=0

We add all the numbers together, and all the variables

15x^2+440x-2090=0

a = 15; b = 440; c = -2090;
Δ = b2-4ac
Δ = 4402-4·15·(-2090)
Δ = 319000
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:

$\sqrt{\Delta}=\sqrt{319000}=\sqrt{100*3190}=\sqrt{100}*\sqrt{3190}=10\sqrt{3190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(440)-10\sqrt{3190}}{2*15}=\frac{-440-10\sqrt{3190}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(440)+10\sqrt{3190}}{2*15}=\frac{-440+10\sqrt{3190}}{30}
$