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5x^2+13x-6=0
a = 5; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·5·(-6)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*5}=\frac{-30}{10} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*5}=\frac{4}{10} =2/5 $
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