0=-5t2+18t+48

Solution for 0=-5t2+18t+48 equation:

0=-5t^2+18t+48

We move all terms to the left:

0-(-5t^2+18t+48)=0

We add all the numbers together, and all the variables

-(-5t^2+18t+48)=0

We get rid of parentheses

5t^2-18t-48=0

a = 5; b = -18; c = -48;
Δ = b2-4ac
Δ = -182-4·5·(-48)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{321}}{2*5}=\frac{18-2\sqrt{321}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{321}}{2*5}=\frac{18+2\sqrt{321}}{10}
$