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.67x^2+.25x-3=0
a = .67; b = .25; c = -3;
Δ = b2-4ac
Δ = .252-4·.67·(-3)
Δ = 8.1025
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{-(.25)-\sqrt{8.1025}}{2*.67}=\frac{-0.25-\sqrt{8.1025}}{1.34} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.25)+\sqrt{8.1025}}{2*.67}=\frac{-0.25+\sqrt{8.1025}}{1.34} $
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