t^2-16t+16=0

Solution for t^2-16t+16=0 equation:

t^2-16t+16=0

a = 1; b = -16; c = +16;
Δ = b2-4ac
Δ = -162-4·1·16
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{3}}{2*1}=\frac{16-8\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{3}}{2*1}=\frac{16+8\sqrt{3}}{2}
$