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15x^2+2x-0.08=0
a = 15; b = 2; c = -0.08;
Δ = b2-4ac
Δ = 22-4·15·(-0.08)
Δ = 8.8
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{-(2)-\sqrt{8.8}}{2*15}=\frac{-2-\sqrt{8.8}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{8.8}}{2*15}=\frac{-2+\sqrt{8.8}}{30} $
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