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10a^2+13a-3=0
a = 10; b = 13; c = -3;
Δ = b2-4ac
Δ = 132-4·10·(-3)
Δ = 289
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{289}=17$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*10}=\frac{-30}{20} =-1+1/2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*10}=\frac{4}{20} =1/5 $
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