5t2+-3t+-2=0

Solution for 5t2+-3t+-2=0 equation:

5t^2+-3t+-2=0

We add all the numbers together, and all the variables

5t^2-3t=0

a = 5; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·5·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*5}=\frac{0}{10}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*5}=\frac{6}{10}
=3/5
$