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m2+12m+2=11
We move all terms to the left:
m2+12m+2-(11)=0
We add all the numbers together, and all the variables
m^2+12m-9=0
a = 1; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·1·(-9)
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{-(12)-6\sqrt{5}}{2*1}=\frac{-12-6\sqrt{5}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{5}}{2*1}=\frac{-12+6\sqrt{5}}{2} $
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