Answer
$\frac{-7}{6}, -3$
Work Step by Step
Given,
$3+ \frac{1}{2p+4} = \frac{10}{(2p+4)^{2}}$
$3+ \frac{1}{2p+4} = 10 (\frac{1}{(2p+4)})^{2}$
Let $x = \frac{1}{2p+4}$ then the equation becomes
$3+x=10x^{2}$
$10x^{2}-x-3=0$
By factoring,
$10x^{2}+5x-6x-3=0$
$5x(2x+1)-3(2x+1)=0$
$(5x-3)(2x+1)=0$
$5x-3=0$
$5x=3$
$x=\frac{3}{5}$
$2x+1=0$
$2x=-1$
$x=\frac{-1}{2}$
Since $x = \frac{1}{2p+4}$ we have,
$x = \frac{1}{2p+4}$
$\frac{3}{5} = \frac{1}{2p+4}$
$3(2p+4)=5$
$6p+12=5$
$6p=5-12$
$6p=-7$
$p=\frac{-7}{6}$
$x = \frac{1}{2p+4}$
$\frac{-1}{2} = \frac{1}{2p+4}$
$-2p-4=2$
$-2p=2+4$
$-2p=6$
$p=-3$
Both $\frac{-7}{6}, -3$ satisfy the given equation, so the solutions are $\frac{-7}{6}, -3$
