15^2=q(2q+16)

Solution for 15^2=q(2q+16) equation:

15^2=q(2q+16)

We move all terms to the left:

15^2-(q(2q+16))=0

We add all the numbers together, and all the variables

-(q(2q+16))+225=0


We calculate terms in parentheses: -(q(2q+16)), so:

q(2q+16)

We multiply parentheses

2q^2+16q

Back to the equation:

-(2q^2+16q)


We get rid of parentheses

-2q^2-16q+225=0

a = -2; b = -16; c = +225;
Δ = b2-4ac
Δ = -162-4·(-2)·225
Δ = 2056
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:

$\sqrt{\Delta}=\sqrt{2056}=\sqrt{4*514}=\sqrt{4}*\sqrt{514}=2\sqrt{514}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{514}}{2*-2}=\frac{16-2\sqrt{514}}{-4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{514}}{2*-2}=\frac{16+2\sqrt{514}}{-4}
$