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5x^2+19x-15=0
a = 5; b = 19; c = -15;
Δ = b2-4ac
Δ = 192-4·5·(-15)
Δ = 661
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{661}}{2*5}=\frac{-19-\sqrt{661}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{661}}{2*5}=\frac{-19+\sqrt{661}}{10} $
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