6t+-2t^2=0

Solution for 6t+-2t^2=0 equation:

6t+-2t^2=0

We add all the numbers together, and all the variables

-2t^2+6t=0

a = -2; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-2)·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*-2}=\frac{-12}{-4}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-2}=\frac{0}{-4}
=0
$