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-0.32x^2+0.4x+-1=0
We add all the numbers together, and all the variables
-0.32x^2+0.4x=0
a = -0.32; b = 0.4; c = 0;
Δ = b2-4ac
Δ = 0.42-4·(-0.32)·0
Δ = 0.16
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{-(0.4)-\sqrt{0.16}}{2*-0.32}=\frac{-0.4-\sqrt{0.16}}{-0.64} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.16}}{2*-0.32}=\frac{-0.4+\sqrt{0.16}}{-0.64} $
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