## Algebra: A Combined Approach (4th Edition)

$p=1$ and $p=\dfrac{5}{7}$
$7p(p-2)+2(p+4)=3$ Evaluate the products present on the left side of the equation: $7p^{2}-14p+2p+8=3$ Take the $3$ to the left side of the equation: $7p^{2}-14p+2p+8-3=0$ Simplify the equation by combining like terms: $7p^{2}-12p+5=0$ Solve this equation using the quadratic formula. This formula is $p=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=7$, $b=-12$ and $c=5$ Substitute: $p=\dfrac{-(-12)\pm\sqrt{(-12)^{2}-4(7)(5)}}{2(7)}=\dfrac{12\pm\sqrt{144-140}}{14}=...$ $...=\dfrac{12\pm\sqrt{4}}{14}=\dfrac{12\pm2}{14}$ Our two solutions are: $p=\dfrac{12+2}{14}=1$ $p=\dfrac{12-2}{14}=\dfrac{5}{7}$