(3(t)^2)-(24t)+(36)=0

Solution for (3(t)^2)-(24t)+(36)=0 equation:

(3(t)^2)-(24t)+(36)=0

determiningTheFunctionDomain
3t^2-24t+36=0

a = 3; b = -24; c = +36;
Δ = b2-4ac
Δ = -242-4·3·36
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12}{2*3}=\frac{12}{6}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12}{2*3}=\frac{36}{6}
=6
$