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m2+8m=-6
We move all terms to the left:
m2+8m-(-6)=0
We add all the numbers together, and all the variables
m^2+8m+6=0
a = 1; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·1·6
Δ = 40
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{10}}{2*1}=\frac{-8-2\sqrt{10}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*1}=\frac{-8+2\sqrt{10}}{2} $
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