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3m^2+44m=0
a = 3; b = 44; c = 0;
Δ = b2-4ac
Δ = 442-4·3·0
Δ = 1936
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{1936}=44$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-44}{2*3}=\frac{-88}{6} =-14+2/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+44}{2*3}=\frac{0}{6} =0 $
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