0=-8t^2+10t+3

Solution for 0=-8t^2+10t+3 equation:

0=-8t^2+10t+3

We move all terms to the left:

0-(-8t^2+10t+3)=0

We add all the numbers together, and all the variables

-(-8t^2+10t+3)=0

We get rid of parentheses

8t^2-10t-3=0

a = 8; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·8·(-3)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*8}=\frac{-4}{16}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*8}=\frac{24}{16}
=1+1/2
$