(2x-20)(3x)=35

Solution for (2x-20)(3x)=35 equation:

(2x-20)(3x)=35

We move all terms to the left:

(2x-20)(3x)-(35)=0

We multiply parentheses

6x^2-60x-35=0

a = 6; b = -60; c = -35;
Δ = b2-4ac
Δ = -602-4·6·(-35)
Δ = 4440
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:

$\sqrt{\Delta}=\sqrt{4440}=\sqrt{4*1110}=\sqrt{4}*\sqrt{1110}=2\sqrt{1110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-2\sqrt{1110}}{2*6}=\frac{60-2\sqrt{1110}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+2\sqrt{1110}}{2*6}=\frac{60+2\sqrt{1110}}{12}
$