2x(3x+20)=28

Solution for 2x(3x+20)=28 equation:

2x(3x+20)=28

We move all terms to the left:

2x(3x+20)-(28)=0

We multiply parentheses

6x^2+40x-28=0

a = 6; b = 40; c = -28;
Δ = b2-4ac
Δ = 402-4·6·(-28)
Δ = 2272
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:

$\sqrt{\Delta}=\sqrt{2272}=\sqrt{16*142}=\sqrt{16}*\sqrt{142}=4\sqrt{142}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{142}}{2*6}=\frac{-40-4\sqrt{142}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{142}}{2*6}=\frac{-40+4\sqrt{142}}{12}
$