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3m^2+31m=-56
We move all terms to the left:
3m^2+31m-(-56)=0
We add all the numbers together, and all the variables
3m^2+31m+56=0
a = 3; b = 31; c = +56;
Δ = b2-4ac
Δ = 312-4·3·56
Δ = 289
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{289}=17$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-17}{2*3}=\frac{-48}{6} =-8 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+17}{2*3}=\frac{-14}{6} =-2+1/3 $
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