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.2x^2-1.6x+.2=0
We add all the numbers together, and all the variables
.2x^2-1.6x+0.2=0
a = .2; b = -1.6; c = +0.2;
Δ = b2-4ac
Δ = -1.62-4·.2·0.2
Δ = 2.4
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{-(-1.6)-\sqrt{2.4}}{2*.2}=\frac{1.6-\sqrt{2.4}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.6)+\sqrt{2.4}}{2*.2}=\frac{1.6+\sqrt{2.4}}{0.4} $
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