M(x)=x^2+40x-800

Solution for M(x)=x^2+40x-800 equation:

(M)=M^2+40M-800

We move all terms to the left:

(M)-(M^2+40M-800)=0

We get rid of parentheses

-M^2+M-40M+800=0

We add all the numbers together, and all the variables

-1M^2-39M+800=0

a = -1; b = -39; c = +800;
Δ = b2-4ac
Δ = -392-4·(-1)·800
Δ = 4721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-\sqrt{4721}}{2*-1}=\frac{39-\sqrt{4721}}{-2}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{4721}}{2*-1}=\frac{39+\sqrt{4721}}{-2}
$