16t^2-56t+13=0

Solution for 16t^2-56t+13=0 equation:

16t^2-56t+13=0

a = 16; b = -56; c = +13;
Δ = b2-4ac
Δ = -562-4·16·13
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-48}{2*16}=\frac{8}{32}
=1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+48}{2*16}=\frac{104}{32}
=3+1/4
$