9^3=14q^2+49q=0

Solution for 9^3=14q^2+49q=0 equation:

9^3=14q^2+49q=0

We move all terms to the left:

9^3-(14q^2+49q)=0

We add all the numbers together, and all the variables

-(14q^2+49q)+729=0

We get rid of parentheses

-14q^2-49q+729=0

a = -14; b = -49; c = +729;
Δ = b2-4ac
Δ = -492-4·(-14)·729
Δ = 43225
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:

$\sqrt{\Delta}=\sqrt{43225}=\sqrt{25*1729}=\sqrt{25}*\sqrt{1729}=5\sqrt{1729}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-5\sqrt{1729}}{2*-14}=\frac{49-5\sqrt{1729}}{-28}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+5\sqrt{1729}}{2*-14}=\frac{49+5\sqrt{1729}}{-28}
$