(5x)(3x+28)=120

Solution for (5x)(3x+28)=120 equation:

(5x)(3x+28)=120

We move all terms to the left:

(5x)(3x+28)-(120)=0

We multiply parentheses

15x^2+140x-120=0

a = 15; b = 140; c = -120;
Δ = b2-4ac
Δ = 1402-4·15·(-120)
Δ = 26800
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:

$\sqrt{\Delta}=\sqrt{26800}=\sqrt{400*67}=\sqrt{400}*\sqrt{67}=20\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-20\sqrt{67}}{2*15}=\frac{-140-20\sqrt{67}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+20\sqrt{67}}{2*15}=\frac{-140+20\sqrt{67}}{30}
$