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15x^2+23x-25=0
a = 15; b = 23; c = -25;
Δ = b2-4ac
Δ = 232-4·15·(-25)
Δ = 2029
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{-(23)-\sqrt{2029}}{2*15}=\frac{-23-\sqrt{2029}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{2029}}{2*15}=\frac{-23+\sqrt{2029}}{30} $
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