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15a^2+13a-5=0
a = 15; b = 13; c = -5;
Δ = b2-4ac
Δ = 132-4·15·(-5)
Δ = 469
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{469}}{2*15}=\frac{-13-\sqrt{469}}{30} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{469}}{2*15}=\frac{-13+\sqrt{469}}{30} $
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