M1=40-4xM2=50-8x

Solution for M1=40-4xM2=50-8x equation:

1=40-4M^2=50-8M

We move all terms to the left:

1-(40-4M^2)=0

We get rid of parentheses

4M^2-40+1=0

We add all the numbers together, and all the variables

4M^2-39=0

a = 4; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·4·(-39)
Δ = 624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{39}}{2*4}=\frac{0-4\sqrt{39}}{8}
=-\frac{4\sqrt{39}}{8}
=-\frac{\sqrt{39}}{2}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{39}}{2*4}=\frac{0+4\sqrt{39}}{8}
=\frac{4\sqrt{39}}{8}
=\frac{\sqrt{39}}{2}
$