2y-3=y^2-4y-12

Solution for 2y-3=y^2-4y-12 equation:

2y-3=y^2-4y-12

We move all terms to the left:

2y-3-(y^2-4y-12)=0

We get rid of parentheses

-y^2+2y+4y+12-3=0

We add all the numbers together, and all the variables

-1y^2+6y+9=0

a = -1; b = 6; c = +9;
Δ = b2-4ac
Δ = 62-4·(-1)·9
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{2}}{2*-1}=\frac{-6-6\sqrt{2}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{2}}{2*-1}=\frac{-6+6\sqrt{2}}{-2}
$