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15x^2+13x-6=0
a = 15; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·15·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*15}=\frac{-36}{30} =-1+1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*15}=\frac{10}{30} =1/3 $
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