z^2+6z-48=0

Solution for z^2+6z-48=0 equation:

z^2+6z-48=0

a = 1; b = 6; c = -48;
Δ = b2-4ac
Δ = 62-4·1·(-48)
Δ = 228
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{57}}{2*1}=\frac{-6-2\sqrt{57}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{57}}{2*1}=\frac{-6+2\sqrt{57}}{2}
$