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-2.4+.03x+.0001x^2=0
a = .0001; b = .03; c = -2.4;
Δ = b2-4ac
Δ = .032-4·.0001·(-2.4)
Δ = 0.00186
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{-(.03)-\sqrt{0.00186}}{2*.0001}=\frac{-0.03-\sqrt{0.00186}}{0.0002} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.03)+\sqrt{0.00186}}{2*.0001}=\frac{-0.03+\sqrt{0.00186}}{0.0002} $
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