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a^2+13a-12=0
a = 1; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·1·(-12)
Δ = 217
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{217}}{2*1}=\frac{-13-\sqrt{217}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{217}}{2*1}=\frac{-13+\sqrt{217}}{2} $
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