M(x)=2x^2+9x-5

Solution for M(x)=2x^2+9x-5 equation:

(M)=2M^2+9M-5

We move all terms to the left:

(M)-(2M^2+9M-5)=0

We get rid of parentheses

-2M^2+M-9M+5=0

We add all the numbers together, and all the variables

-2M^2-8M+5=0

a = -2; b = -8; c = +5;
Δ = b2-4ac
Δ = -82-4·(-2)·5
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{26}}{2*-2}=\frac{8-2\sqrt{26}}{-4}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{26}}{2*-2}=\frac{8+2\sqrt{26}}{-4}
$