(18=8t+0.5t^2)

Solution for (18=8t+0.5t^2) equation:

(18=8t+0.5t^2)

We move all terms to the left:

(18-(8t+0.5t^2))=0


We calculate terms in parentheses: +(18-(8t+0.5t^2)), so:

18-(8t+0.5t^2)

determiningTheFunctionDomain
-(8t+0.5t^2)+18

We get rid of parentheses

-0.5t^2-8t+18

Back to the equation:

+(-0.5t^2-8t+18)


We get rid of parentheses

-0.5t^2-8t+18=0

a = -0.5; b = -8; c = +18;
Δ = b2-4ac
Δ = -82-4·(-0.5)·18
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-10}{2*-0.5}=\frac{-2}{-1}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+10}{2*-0.5}=\frac{18}{-1}
=-18
$