0=5+(2.7)t-4.9t^2

Solution for 0=5+(2.7)t-4.9t^2 equation:

0=5+(2.7)t-4.9t^2

We move all terms to the left:

0-(5+(2.7)t-4.9t^2)=0

We add all the numbers together, and all the variables

-(5+(2.7)t-4.9t^2)=0


We calculate terms in parentheses: -(5+(2.7)t-4.9t^2), so:

5+(2.7)t-4.9t^2

determiningTheFunctionDomain
-4.9t^2+(2.7)t+5

We multiply parentheses

-4.9t^2+2.7t+5

Back to the equation:

-(-4.9t^2+2.7t+5)


We get rid of parentheses

4.9t^2-2.7t-5=0

a = 4.9; b = -2.7; c = -5;
Δ = b2-4ac
Δ = -2.72-4·4.9·(-5)
Δ = 105.29
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{-(-2.7)-\sqrt{105.29}}{2*4.9}=\frac{2.7-\sqrt{105.29}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.7)+\sqrt{105.29}}{2*4.9}=\frac{2.7+\sqrt{105.29}}{9.8}
$