16(t^2-4t+3)=0

Solution for 16(t^2-4t+3)=0 equation:

16(t^2-4t+3)=0

We multiply parentheses

16t^2-64t+48=0

a = 16; b = -64; c = +48;
Δ = b2-4ac
Δ = -642-4·16·48
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-32}{2*16}=\frac{32}{32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+32}{2*16}=\frac{96}{32}
=3
$