z^2+14z-15=0

Solution for z^2+14z-15=0 equation:

z^2+14z-15=0

a = 1; b = 14; c = -15;
Δ = b2-4ac
Δ = 142-4·1·(-15)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-16}{2*1}=\frac{-30}{2}
=-15
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*1}=\frac{2}{2}
=1
$