(z+4)(z+5)=10

Solution for (z+4)(z+5)=10 equation:

(z+4)(z+5)=10

We move all terms to the left:

(z+4)(z+5)-(10)=0

We multiply parentheses ..

(+z^2+5z+4z+20)-10=0

We get rid of parentheses

z^2+5z+4z+20-10=0

We add all the numbers together, and all the variables

z^2+9z+10=0

a = 1; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·1·10
Δ = 41
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{-(9)-\sqrt{41}}{2*1}=\frac{-9-\sqrt{41}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{41}}{2*1}=\frac{-9+\sqrt{41}}{2}
$