2=-16t^2+147t+0

Solution for 2=-16t^2+147t+0 equation:

2=-16t^2+147t+0

We move all terms to the left:

2-(-16t^2+147t+0)=0

We get rid of parentheses

16t^2-147t-0+2=0

We add all the numbers together, and all the variables

16t^2-147t+2=0

a = 16; b = -147; c = +2;
Δ = b2-4ac
Δ = -1472-4·16·2
Δ = 21481
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{-(-147)-\sqrt{21481}}{2*16}=\frac{147-\sqrt{21481}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-147)+\sqrt{21481}}{2*16}=\frac{147+\sqrt{21481}}{32}
$