0=-16t^2+150t+48

Solution for 0=-16t^2+150t+48 equation:

0=-16t^2+150t+48

We move all terms to the left:

0-(-16t^2+150t+48)=0

We add all the numbers together, and all the variables

-(-16t^2+150t+48)=0

We get rid of parentheses

16t^2-150t-48=0

a = 16; b = -150; c = -48;
Δ = b2-4ac
Δ = -1502-4·16·(-48)
Δ = 25572
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:

$\sqrt{\Delta}=\sqrt{25572}=\sqrt{4*6393}=\sqrt{4}*\sqrt{6393}=2\sqrt{6393}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-2\sqrt{6393}}{2*16}=\frac{150-2\sqrt{6393}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+2\sqrt{6393}}{2*16}=\frac{150+2\sqrt{6393}}{32}
$