t2+16t=64

Solution for t2+16t=64 equation:

t2+16t=64

We move all terms to the left:

t2+16t-(64)=0

We add all the numbers together, and all the variables

t^2+16t-64=0

a = 1; b = 16; c = -64;
Δ = b2-4ac
Δ = 162-4·1·(-64)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*1}=\frac{-16-16\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*1}=\frac{-16+16\sqrt{2}}{2}
$