0=-5t2-4t+3

Solution for 0=-5t2-4t+3 equation:

0=-5t^2-4t+3

We move all terms to the left:

0-(-5t^2-4t+3)=0

We add all the numbers together, and all the variables

-(-5t^2-4t+3)=0

We get rid of parentheses

5t^2+4t-3=0

a = 5; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·5·(-3)
Δ = 76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{19}}{2*5}=\frac{-4-2\sqrt{19}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{19}}{2*5}=\frac{-4+2\sqrt{19}}{10}
$