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15x^2-3x-10=0
a = 15; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·15·(-10)
Δ = 609
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{-(-3)-\sqrt{609}}{2*15}=\frac{3-\sqrt{609}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{609}}{2*15}=\frac{3+\sqrt{609}}{30} $
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