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3x^2+.167x-3.167=0
a = 3; b = .167; c = -3.167;
Δ = b2-4ac
Δ = .1672-4·3·(-3.167)
Δ = 38.031889
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{-(.167)-\sqrt{38.031889}}{2*3}=\frac{-0.167-\sqrt{38.031889}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.167)+\sqrt{38.031889}}{2*3}=\frac{-0.167+\sqrt{38.031889}}{6} $
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