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-.02x^2+0.4x+0.8=0
We add all the numbers together, and all the variables
-0.02x^2+0.4x+0.8=0
a = -0.02; b = 0.4; c = +0.8;
Δ = b2-4ac
Δ = 0.42-4·(-0.02)·0.8
Δ = 0.224
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{0.224}}{2*-0.02}=\frac{-0.4-\sqrt{0.224}}{-0.04} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.224}}{2*-0.02}=\frac{-0.4+\sqrt{0.224}}{-0.04} $
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