16t^2-68t+13=60t

Solution for 16t^2-68t+13=60t equation:

16t^2-68t+13=60t

We move all terms to the left:

16t^2-68t+13-(60t)=0

We add all the numbers together, and all the variables

16t^2-128t+13=0

a = 16; b = -128; c = +13;
Δ = b2-4ac
Δ = -1282-4·16·13
Δ = 15552
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:

$\sqrt{\Delta}=\sqrt{15552}=\sqrt{5184*3}=\sqrt{5184}*\sqrt{3}=72\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-72\sqrt{3}}{2*16}=\frac{128-72\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+72\sqrt{3}}{2*16}=\frac{128+72\sqrt{3}}{32}
$