y^2+4y-48=0

Solution for y^2+4y-48=0 equation:

y^2+4y-48=0

a = 1; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·1·(-48)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{13}}{2*1}=\frac{-4-4\sqrt{13}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{13}}{2*1}=\frac{-4+4\sqrt{13}}{2}
$