2y(3y+8)=40

Solution for 2y(3y+8)=40 equation:

2y(3y+8)=40

We move all terms to the left:

2y(3y+8)-(40)=0

We multiply parentheses

6y^2+16y-40=0

a = 6; b = 16; c = -40;
Δ = b2-4ac
Δ = 162-4·6·(-40)
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{19}}{2*6}=\frac{-16-8\sqrt{19}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{19}}{2*6}=\frac{-16+8\sqrt{19}}{12}
$