16t^2-55t+25=0

Solution for 16t^2-55t+25=0 equation:

16t^2-55t+25=0

a = 16; b = -55; c = +25;
Δ = b2-4ac
Δ = -552-4·16·25
Δ = 1425
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:

$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-5\sqrt{57}}{2*16}=\frac{55-5\sqrt{57}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+5\sqrt{57}}{2*16}=\frac{55+5\sqrt{57}}{32}
$