16=4(q+1)q=

Solution for 16=4(q+1)q= equation:

16=4(q+1)q=

We move all terms to the left:

16-(4(q+1)q)=0


We calculate terms in parentheses: -(4(q+1)q), so:

4(q+1)q

We multiply parentheses

4q^2+4q

Back to the equation:

-(4q^2+4q)


We get rid of parentheses

-4q^2-4q+16=0

a = -4; b = -4; c = +16;
Δ = b2-4ac
Δ = -42-4·(-4)·16
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{17}}{2*-4}=\frac{4-4\sqrt{17}}{-8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{17}}{2*-4}=\frac{4+4\sqrt{17}}{-8}
$