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6x^2-42x+40=0

a = 6; b = -42; c = +40;

Δ = b^{2}-4ac

Δ = -42^{2}-4·6·40

Δ = 804

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{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{201}}{2*6}=\frac{42-2\sqrt{201}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{201}}{2*6}=\frac{42+2\sqrt{201}}{12} $

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