Solve the equation (7/20)x + (x - 2)/15 + (1/12)x^2 = (1/12)(x + 3)^2 - (1/20)(2x + 3).

Step-by-step solution

1. 1

   Expand the right side. $(x+3)^2=x^2+6x+9$, so $\frac{1}{12}(x+3)^2=\frac{1}{12}x^2+\frac12x+\frac34$. Also $-\frac{1}{20}(2x+3)=-\frac{1}{10}x-\frac{3}{20}$.

2. 2

   Collect the right side. $\frac12x-\frac{1}{10}x=\frac{5-1}{10}x=\frac25x$ and $\frac34-\frac{3}{20}=\frac{15-3}{20}=\frac35$, so the right side is $\frac{1}{12}x^2+\frac25x+\frac35$.

3. 3

   Collect the left side. Split $\frac{x-2}{15}=\frac{x}{15}-\frac{2}{15}$. Then $\frac{7}{20}x+\frac{1}{15}x=\frac{21}{60}x+\frac{4}{60}x=\frac{25}{60}x=\frac{5}{12}x$, giving $\frac{1}{12}x^2+\frac{5}{12}x-\frac{2}{15}$.

4. 4

   Cancel the quadratic term. Both sides contain exactly $\frac{1}{12}x^2$, so subtracting it turns a seemingly quadratic equation into the linear one $\frac{5}{12}x-\frac{2}{15}=\frac25x+\frac35$.

5. 5

   Gather $x$ on the left and constants on the right. $\frac{5}{12}x-\frac25x=\frac{25}{60}x-\frac{24}{60}x=\frac{1}{60}x$, while $\frac35+\frac{2}{15}=\frac{9}{15}+\frac{2}{15}=\frac{11}{15}$. The equation is $\frac{1}{60}x=\frac{11}{15}$.

6. 6

   Solve and verify. Multiplying by $60$ gives $x=\frac{11}{15}\cdot 60=44$. Substituting $x=44$ makes both sides equal $\frac{2693}{15}$, confirming the root.