Answer
$x=-\displaystyle \frac{11}{8}+\frac{\sqrt{41}}{8}$ or $x=-\displaystyle \frac{11}{8}-\frac{\sqrt{41}}{8}$
Work Step by Step
$ 7x(x+2)+5=3x(x+1)\qquad$... use the distributive property: $a(b+c)=ab+ac$.
$ 7x^{2}+14x+5=3x^{2}+3x\qquad$...add $(-3x^{2}-3x)$ to both sides.
$ 7x^{2}+14x+5-3x^{2}-3x=3x^{2}+3x-3x^{2}-3x\qquad$...add like terms.
$ 4x^{2}+11x+5=0\qquad$... solve with the Quadractic formula. $a=4,\ b=11,\ c=5$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $11,\ a$ for $4$ and $c$ for $5$.
$ x=\displaystyle \frac{-11\pm\sqrt{(11)^{2}-4\cdot(4)\cdot 5}}{2\cdot 4}\qquad$... simplify.
$x=\displaystyle \frac{-11\pm\sqrt{121-80}}{8}$
$ x=\displaystyle \frac{-11\pm\sqrt{41}}{8}\qquad$... the symbol $\pm$ indicates two solutions.
$x=-\displaystyle \frac{11}{8}+\frac{\sqrt{41}}{8}$ or $x=-\displaystyle \frac{11}{8}-\frac{\sqrt{41}}{8}$
