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3.2x+.2x^2-5=0
a = .2; b = 3.2; c = -5;
Δ = b2-4ac
Δ = 3.22-4·.2·(-5)
Δ = 14.24
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{-(3.2)-\sqrt{14.24}}{2*.2}=\frac{-3.2-\sqrt{14.24}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.2)+\sqrt{14.24}}{2*.2}=\frac{-3.2+\sqrt{14.24}}{0.4} $
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