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3m^2+29m+56=0
a = 3; b = 29; c = +56;
Δ = b2-4ac
Δ = 292-4·3·56
Δ = 169
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{169}=13$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-13}{2*3}=\frac{-42}{6} =-7 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+13}{2*3}=\frac{-16}{6} =-2+2/3 $
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