3+-4t+2t^2=0

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

3+-4t+2t^2=0

We add all the numbers together, and all the variables

2t^2-4t=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$