z^2+8z-10=0

Solution for z^2+8z-10=0 equation:

z^2+8z-10=0

a = 1; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·1·(-10)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{26}}{2*1}=\frac{-8-2\sqrt{26}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{26}}{2*1}=\frac{-8+2\sqrt{26}}{2}
$