2z+5=z^2-10

Solution for 2z+5=z^2-10 equation:

2z+5=z^2-10

We move all terms to the left:

2z+5-(z^2-10)=0

We get rid of parentheses

-z^2+2z+10+5=0

We add all the numbers together, and all the variables

-1z^2+2z+15=0

a = -1; b = 2; c = +15;
Δ = b2-4ac
Δ = 22-4·(-1)·15
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-1}=\frac{-10}{-2}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-1}=\frac{6}{-2}
=-3
$