3t2+-6t+1=0

Solution for 3t2+-6t+1=0 equation:

3t^2+-6t+1=0

We add all the numbers together, and all the variables

3t^2-6t=0

a = 3; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·3·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*3}=\frac{0}{6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*3}=\frac{12}{6}
=2
$