6t^2=17+14=0

Solution for 6t^2=17+14=0 equation:

6t^2=17+14=0

We move all terms to the left:

6t^2-(17+14)=0

We add all the numbers together, and all the variables

6t^2-31=0

a = 6; b = 0; c = -31;
Δ = b2-4ac
Δ = 02-4·6·(-31)
Δ = 744
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:

$\sqrt{\Delta}=\sqrt{744}=\sqrt{4*186}=\sqrt{4}*\sqrt{186}=2\sqrt{186}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{186}}{2*6}=\frac{0-2\sqrt{186}}{12}
=-\frac{2\sqrt{186}}{12}
=-\frac{\sqrt{186}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{186}}{2*6}=\frac{0+2\sqrt{186}}{12}
=\frac{2\sqrt{186}}{12}
=\frac{\sqrt{186}}{6}
$