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m2+33m=0
We add all the numbers together, and all the variables
m^2+33m=0
a = 1; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·1·0
Δ = 1089
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{1089}=33$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*1}=\frac{-66}{2} =-33 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*1}=\frac{0}{2} =0 $
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