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-0.3x+-0.2x^2=0.0
We add all the numbers together, and all the variables
-0.2x^2-0.3x=0
a = -0.2; b = -0.3; c = 0;
Δ = b2-4ac
Δ = -0.32-4·(-0.2)·0
Δ = 0.09
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{-(-0.3)-\sqrt{0.09}}{2*-0.2}=\frac{0.3-\sqrt{0.09}}{-0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.3)+\sqrt{0.09}}{2*-0.2}=\frac{0.3+\sqrt{0.09}}{-0.4} $
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