0=-4.9t^2+150t+350

Solution for 0=-4.9t^2+150t+350 equation:

0=-4.9t^2+150t+350

We move all terms to the left:

0-(-4.9t^2+150t+350)=0

We add all the numbers together, and all the variables

-(-4.9t^2+150t+350)=0

We get rid of parentheses

4.9t^2-150t-350=0

a = 4.9; b = -150; c = -350;
Δ = b2-4ac
Δ = -1502-4·4.9·(-350)
Δ = 29360
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:

$\sqrt{\Delta}=\sqrt{29360}=\sqrt{16*1835}=\sqrt{16}*\sqrt{1835}=4\sqrt{1835}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-4\sqrt{1835}}{2*4.9}=\frac{150-4\sqrt{1835}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+4\sqrt{1835}}{2*4.9}=\frac{150+4\sqrt{1835}}{9.8}
$