6t^2+36t+1=0

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Solution for 6t^2+36t+1=0 equation:



6t^2+36t+1=0
a = 6; b = 36; c = +1;
Δ = b2-4ac
Δ = 362-4·6·1
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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