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6x^2-42x+40=0
a = 6; b = -42; c = +40;
Δ = b2-4ac
Δ = -422-4·6·40
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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