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a^2+13a-90=0
a = 1; b = 13; c = -90;
Δ = b2-4ac
Δ = 132-4·1·(-90)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*1}=\frac{-36}{2} =-18 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*1}=\frac{10}{2} =5 $
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