3m+5m=m^2-7m-18

Solution for 3m+5m=m^2-7m-18 equation:

3m+5m=m^2-7m-18

We move all terms to the left:

3m+5m-(m^2-7m-18)=0

We add all the numbers together, and all the variables

8m-(m^2-7m-18)=0

We get rid of parentheses

-m^2+8m+7m+18=0

We add all the numbers together, and all the variables

-1m^2+15m+18=0

a = -1; b = 15; c = +18;
Δ = b2-4ac
Δ = 152-4·(-1)·18
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{33}}{2*-1}=\frac{-15-3\sqrt{33}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{33}}{2*-1}=\frac{-15+3\sqrt{33}}{-2}
$