4m^2+8=-18m

Solution for 4m^2+8=-18m equation:

4m^2+8=-18m

We move all terms to the left:

4m^2+8-(-18m)=0

We get rid of parentheses

4m^2+18m+8=0

a = 4; b = 18; c = +8;
Δ = b2-4ac
Δ = 182-4·4·8
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-14}{2*4}=\frac{-32}{8}
=-4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+14}{2*4}=\frac{-4}{8}
=-1/2
$