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m2+8m=48
We move all terms to the left:
m2+8m-(48)=0
We add all the numbers together, and all the variables
m^2+8m-48=0
a = 1; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·1·(-48)
Δ = 256
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{256}=16$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*1}=\frac{-24}{2} =-12 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*1}=\frac{8}{2} =4 $
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