6t^2-18t+2=0

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Solution for 6t^2-18t+2=0 equation:



6t^2-18t+2=0
a = 6; b = -18; c = +2;
Δ = b2-4ac
Δ = -182-4·6·2
Δ = 276
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{69}}{2*6}=\frac{18-2\sqrt{69}}{12} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{69}}{2*6}=\frac{18+2\sqrt{69}}{12} $

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