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.05x^2-x+4=0
We add all the numbers together, and all the variables
.05x^2-1x+4=0
a = .05; b = -1; c = +4;
Δ = b2-4ac
Δ = -12-4·.05·4
Δ = 0.2
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)-\sqrt{0.2}}{2*.05}=\frac{1-\sqrt{0.2}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{0.2}}{2*.05}=\frac{1+\sqrt{0.2}}{0.1} $
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