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0.1x^2-0.15x-0.24=0
a = 0.1; b = -0.15; c = -0.24;
Δ = b2-4ac
Δ = -0.152-4·0.1·(-0.24)
Δ = 0.1185
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{-(-0.15)-\sqrt{0.1185}}{2*0.1}=\frac{0.15-\sqrt{0.1185}}{0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.15)+\sqrt{0.1185}}{2*0.1}=\frac{0.15+\sqrt{0.1185}}{0.2} $
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