-36-2a=-a^2+4-5a

Solution for -36-2a=-a^2+4-5a equation:

-36-2a=-a^2+4-5a

We move all terms to the left:

-36-2a-(-a^2+4-5a)=0

We get rid of parentheses

a^2+5a-2a-4-36=0

We add all the numbers together, and all the variables

a^2+3a-40=0

a = 1; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*1}=\frac{-16}{2}
=-8
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*1}=\frac{10}{2}
=5
$