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y^2+y-132=0
a = 1; b = 1; c = -132;
Δ = b2-4ac
Δ = 12-4·1·(-132)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-23}{2*1}=\frac{-24}{2} =-12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+23}{2*1}=\frac{22}{2} =11 $
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