82-202t+16t^2=0

Solution for 82-202t+16t^2=0 equation:

82-202t+16t^2=0

a = 16; b = -202; c = +82;
Δ = b2-4ac
Δ = -2022-4·16·82
Δ = 35556
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:

$\sqrt{\Delta}=\sqrt{35556}=\sqrt{4*8889}=\sqrt{4}*\sqrt{8889}=2\sqrt{8889}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-202)-2\sqrt{8889}}{2*16}=\frac{202-2\sqrt{8889}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-202)+2\sqrt{8889}}{2*16}=\frac{202+2\sqrt{8889}}{32}
$