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4.8+2x+.2x^2=0
a = .2; b = 2; c = +4.8;
Δ = b2-4ac
Δ = 22-4·.2·4.8
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{-(2)-\sqrt{0.16}}{2*.2}=\frac{-2-\sqrt{0.16}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{0.16}}{2*.2}=\frac{-2+\sqrt{0.16}}{0.4} $
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