0=-5t2+2t+3

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

5t^2-2t-3=0

a = 5; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·5·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*5}=\frac{-6}{10}
=-3/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*5}=\frac{10}{10}
=1
$