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15x^2+2x-8=0
a = 15; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·15·(-8)
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*15}=\frac{-24}{30} =-4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*15}=\frac{20}{30} =2/3 $
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