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y^2+3y-180=0
a = 1; b = 3; c = -180;
Δ = b2-4ac
Δ = 32-4·1·(-180)
Δ = 729
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{729}=27$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*1}=\frac{-30}{2} =-15 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*1}=\frac{24}{2} =12 $
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