2m(m+3)=18

Solution for 2m(m+3)=18 equation:

2m(m+3)=18

We move all terms to the left:

2m(m+3)-(18)=0

We multiply parentheses

2m^2+6m-18=0

a = 2; b = 6; c = -18;
Δ = b2-4ac
Δ = 62-4·2·(-18)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{5}}{2*2}=\frac{-6-6\sqrt{5}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{5}}{2*2}=\frac{-6+6\sqrt{5}}{4}
$