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4m^2+2m-132=0
a = 4; b = 2; c = -132;
Δ = b2-4ac
Δ = 22-4·4·(-132)
Δ = 2116
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{2116}=46$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-46}{2*4}=\frac{-48}{8} =-6 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+46}{2*4}=\frac{44}{8} =5+1/2 $
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