m^2+12m+18=7

Solution for m^2+12m+18=7 equation:

m^2+12m+18=7

We move all terms to the left:

m^2+12m+18-(7)=0

We add all the numbers together, and all the variables

m^2+12m+11=0

a = 1; b = 12; c = +11;
Δ = b2-4ac
Δ = 122-4·1·11
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-10}{2*1}=\frac{-22}{2}
=-11
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+10}{2*1}=\frac{-2}{2}
=-1
$