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3x^2-18x+2.5=0
a = 3; b = -18; c = +2.5;
Δ = b2-4ac
Δ = -182-4·3·2.5
Δ = 294
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{294}=\sqrt{49*6}=\sqrt{49}*\sqrt{6}=7\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-7\sqrt{6}}{2*3}=\frac{18-7\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+7\sqrt{6}}{2*3}=\frac{18+7\sqrt{6}}{6} $
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