(4y-3)y=3

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

(4y-3)y=3

We move all terms to the left:

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

We multiply parentheses

4y^2-3y-3=0

a = 4; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·4·(-3)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{57}}{2*4}=\frac{3-\sqrt{57}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*4}=\frac{3+\sqrt{57}}{8}
$