4y^2-4=15y

Solution for 4y^2-4=15y equation:

4y^2-4=15y

We move all terms to the left:

4y^2-4-(15y)=0

a = 4; b = -15; c = -4;
Δ = b2-4ac
Δ = -152-4·4·(-4)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-17}{2*4}=\frac{-2}{8}
=-1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+17}{2*4}=\frac{32}{8}
=4
$