2y+5=y^2+4y-13

Solution for 2y+5=y^2+4y-13 equation:

2y+5=y^2+4y-13

We move all terms to the left:

2y+5-(y^2+4y-13)=0

We get rid of parentheses

-y^2+2y-4y+13+5=0

We add all the numbers together, and all the variables

-1y^2-2y+18=0

a = -1; b = -2; c = +18;
Δ = b2-4ac
Δ = -22-4·(-1)·18
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{19}}{2*-1}=\frac{2-2\sqrt{19}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{19}}{2*-1}=\frac{2+2\sqrt{19}}{-2}
$