4t^2=68

Solution for 4t^2=68 equation:

4t^2=68

We move all terms to the left:

4t^2-(68)=0

a = 4; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·4·(-68)
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{17}}{2*4}=\frac{0-8\sqrt{17}}{8}
=-\frac{8\sqrt{17}}{8}
=-\sqrt{17}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{17}}{2*4}=\frac{0+8\sqrt{17}}{8}
=\frac{8\sqrt{17}}{8}
=\sqrt{17}
$