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15x^2+9x-8=0
a = 15; b = 9; c = -8;
Δ = b2-4ac
Δ = 92-4·15·(-8)
Δ = 561
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{-(9)-\sqrt{561}}{2*15}=\frac{-9-\sqrt{561}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{561}}{2*15}=\frac{-9+\sqrt{561}}{30} $
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