y^2+13y=48

Solution for y^2+13y=48 equation:

y^2+13y=48

We move all terms to the left:

y^2+13y-(48)=0

a = 1; b = 13; c = -48;
Δ = b2-4ac
Δ = 132-4·1·(-48)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*1}=\frac{-32}{2}
=-16
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*1}=\frac{6}{2}
=3
$