(2y)(3y+10)=70

Solution for (2y)(3y+10)=70 equation:

(2y)(3y+10)=70

We move all terms to the left:

(2y)(3y+10)-(70)=0

We multiply parentheses

6y^2+20y-70=0

a = 6; b = 20; c = -70;
Δ = b2-4ac
Δ = 202-4·6·(-70)
Δ = 2080
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{130}}{2*6}=\frac{-20-4\sqrt{130}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{130}}{2*6}=\frac{-20+4\sqrt{130}}{12}
$