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.02x^2+.4x-6=0
a = .02; b = .4; c = -6;
Δ = b2-4ac
Δ = .42-4·.02·(-6)
Δ = 0.64
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{0.64}}{2*.02}=\frac{-0.4-\sqrt{0.64}}{0.04} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.4)+\sqrt{0.64}}{2*.02}=\frac{-0.4+\sqrt{0.64}}{0.04} $
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