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2x^2+13x-132=0
a = 2; b = 13; c = -132;
Δ = b2-4ac
Δ = 132-4·2·(-132)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-35}{2*2}=\frac{-48}{4} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+35}{2*2}=\frac{22}{4} =5+1/2 $
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