16t^2-55t+26=0

Solution for 16t^2-55t+26=0 equation:

16t^2-55t+26=0

a = 16; b = -55; c = +26;
Δ = b2-4ac
Δ = -552-4·16·26
Δ = 1361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-\sqrt{1361}}{2*16}=\frac{55-\sqrt{1361}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+\sqrt{1361}}{2*16}=\frac{55+\sqrt{1361}}{32}
$