54=100-24q+3q^2

Solution for 54=100-24q+3q^2 equation:

54=100-24q+3q^2

We move all terms to the left:

54-(100-24q+3q^2)=0

We get rid of parentheses

-3q^2+24q-100+54=0

We add all the numbers together, and all the variables

-3q^2+24q-46=0

a = -3; b = 24; c = -46;
Δ = b2-4ac
Δ = 242-4·(-3)·(-46)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{6}}{2*-3}=\frac{-24-2\sqrt{6}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{6}}{2*-3}=\frac{-24+2\sqrt{6}}{-6}
$