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m^2+12m=-13
We move all terms to the left:
m^2+12m-(-13)=0
We add all the numbers together, and all the variables
m^2+12m+13=0
a = 1; b = 12; c = +13;
Δ = b2-4ac
Δ = 122-4·1·13
Δ = 92
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{23}}{2*1}=\frac{-12-2\sqrt{23}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{23}}{2*1}=\frac{-12+2\sqrt{23}}{2} $
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