4(2y-3)y=5

Solution for 4(2y-3)y=5 equation:

4(2y-3)y=5

We move all terms to the left:

4(2y-3)y-(5)=0

We multiply parentheses

8y^2-12y-5=0

a = 8; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·8·(-5)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{19}}{2*8}=\frac{12-4\sqrt{19}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{19}}{2*8}=\frac{12+4\sqrt{19}}{16}
$