-200=(18.13)(t)-4.9(t)^2

Solution for -200=(18.13)(t)-4.9(t)^2 equation:

-200=(18.13)(t)-4.9(t)^2

We move all terms to the left:

-200-((18.13)(t)-4.9(t)^2)=0


We calculate terms in parentheses: -((18.13)t-4.9t^2), so:

(18.13)t-4.9t^2

determiningTheFunctionDomain
-4.9t^2+(18.13)t

We multiply parentheses

-4.9t^2+18.13t

Back to the equation:

-(-4.9t^2+18.13t)


We get rid of parentheses

4.9t^2-18.13t-200=0

a = 4.9; b = -18.13; c = -200;
Δ = b2-4ac
Δ = -18.132-4·4.9·(-200)
Δ = 4248.6969
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{-(-18.13)-\sqrt{4248.6969}}{2*4.9}=\frac{18.13-\sqrt{4248.6969}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18.13)+\sqrt{4248.6969}}{2*4.9}=\frac{18.13+\sqrt{4248.6969}}{9.8}
$