m=4m^2-21m-18

Solution for m=4m^2-21m-18 equation:

m=4m^2-21m-18

We move all terms to the left:

m-(4m^2-21m-18)=0

We get rid of parentheses

-4m^2+m+21m+18=0

We add all the numbers together, and all the variables

-4m^2+22m+18=0

a = -4; b = 22; c = +18;
Δ = b2-4ac
Δ = 222-4·(-4)·18
Δ = 772
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:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{193}}{2*-4}=\frac{-22-2\sqrt{193}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{193}}{2*-4}=\frac{-22+2\sqrt{193}}{-8}
$