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15x^2+2.4x-0.64=0
a = 15; b = 2.4; c = -0.64;
Δ = b2-4ac
Δ = 2.42-4·15·(-0.64)
Δ = 44.16
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{-(2.4)-\sqrt{44.16}}{2*15}=\frac{-2.4-\sqrt{44.16}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.4)+\sqrt{44.16}}{2*15}=\frac{-2.4+\sqrt{44.16}}{30} $
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