0=-5t^2+15t+0.5

Solution for 0=-5t^2+15t+0.5 equation:

0=-5t^2+15t+0.5

We move all terms to the left:

0-(-5t^2+15t+0.5)=0

We add all the numbers together, and all the variables

-(-5t^2+15t+0.5)=0

We get rid of parentheses

5t^2-15t-0.5=0

a = 5; b = -15; c = -0.5;
Δ = b2-4ac
Δ = -152-4·5·(-0.5)
Δ = 235
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{235}}{2*5}=\frac{15-\sqrt{235}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{235}}{2*5}=\frac{15+\sqrt{235}}{10}
$