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2m^2+5m-133=0
a = 2; b = 5; c = -133;
Δ = b2-4ac
Δ = 52-4·2·(-133)
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-33}{2*2}=\frac{-38}{4} =-9+1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+33}{2*2}=\frac{28}{4} =7 $
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