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m^2+3m=54
We move all terms to the left:
m^2+3m-(54)=0
a = 1; b = 3; c = -54;
Δ = b2-4ac
Δ = 32-4·1·(-54)
Δ = 225
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{225}=15$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*1}=\frac{-18}{2} =-9 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*1}=\frac{12}{2} =6 $
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