z2+7z-8=20

Solution for z2+7z-8=20 equation:

z2+7z-8=20

We move all terms to the left:

z2+7z-8-(20)=0

We add all the numbers together, and all the variables

z^2+7z-28=0

a = 1; b = 7; c = -28;
Δ = b2-4ac
Δ = 72-4·1·(-28)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{161}}{2*1}=\frac{-7-\sqrt{161}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{161}}{2*1}=\frac{-7+\sqrt{161}}{2}
$