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3x^2-18x+1.4=0
a = 3; b = -18; c = +1.4;
Δ = b2-4ac
Δ = -182-4·3·1.4
Δ = 307.2
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{-(-18)-\sqrt{307.2}}{2*3}=\frac{18-\sqrt{307.2}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+\sqrt{307.2}}{2*3}=\frac{18+\sqrt{307.2}}{6} $
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