t2=612

Solution for t2=612 equation:

t2=612

We move all terms to the left:

t2-(612)=0

We add all the numbers together, and all the variables

t^2-612=0

a = 1; b = 0; c = -612;
Δ = b2-4ac
Δ = 02-4·1·(-612)
Δ = 2448
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:

$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{17}}{2*1}=\frac{0-12\sqrt{17}}{2}
=-\frac{12\sqrt{17}}{2}
=-6\sqrt{17}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{17}}{2*1}=\frac{0+12\sqrt{17}}{2}
=\frac{12\sqrt{17}}{2}
=6\sqrt{17}
$