0=(4.9t^2)+49t+1.5

Solution for 0=(4.9t^2)+49t+1.5 equation:

0=(4.9t^2)+49t+1.5

We move all terms to the left:

0-((4.9t^2)+49t+1.5)=0

We add all the numbers together, and all the variables

-((4.9t^2)+49t+1.5)=0


We calculate terms in parentheses: -((4.9t^2)+49t+1.5), so:

(4.9t^2)+49t+1.5

We add all the numbers together, and all the variables

49t+(4.9t^2)+1.5

We get rid of parentheses

4.9t^2+49t+1.5

Back to the equation:

-(4.9t^2+49t+1.5)


We get rid of parentheses

-4.9t^2-49t-1.5=0

a = -4.9; b = -49; c = -1.5;
Δ = b2-4ac
Δ = -492-4·(-4.9)·(-1.5)
Δ = 2371.6
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{-(-49)-\sqrt{2371.6}}{2*-4.9}=\frac{49-\sqrt{2371.6}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2371.6}}{2*-4.9}=\frac{49+\sqrt{2371.6}}{-9.8}
$