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6t^2+36t+1=0

a = 6; b = 36; c = +1;

Δ = b^{2}-4ac

Δ = 36^{2}-4·6·1

Δ = 1272

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{1272}=\sqrt{4*318}=\sqrt{4}*\sqrt{318}=2\sqrt{318}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{318}}{2*6}=\frac{-36-2\sqrt{318}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{318}}{2*6}=\frac{-36+2\sqrt{318}}{12} $

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