D(126)=(6t^2)+24(t)

Solution for D(126)=(6t^2)+24(t) equation:

(126)=(6D^2)+24(D)

We move all terms to the left:

(126)-((6D^2)+24(D))=0

We get rid of parentheses

-6D^2-24D+126=0

a = -6; b = -24; c = +126;
Δ = b2-4ac
Δ = -242-4·(-6)·126
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-60}{2*-6}=\frac{-36}{-12}
=+3
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+60}{2*-6}=\frac{84}{-12}
=-7
$