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2x^2+4x-0.16=0
a = 2; b = 4; c = -0.16;
Δ = b2-4ac
Δ = 42-4·2·(-0.16)
Δ = 17.28
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{-(4)-\sqrt{17.28}}{2*2}=\frac{-4-\sqrt{17.28}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{17.28}}{2*2}=\frac{-4+\sqrt{17.28}}{4} $
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