$\displaystyle \frac {4}{a-7} = \frac {-2a}{a+3}$
$\displaystyle \frac{4}
{{a - 7}} = \frac{{ - 2a}}
{{a + 3}}$
$\displaystyle 4\left( {a + 3} \right) = - 2a\left( {a - 7} \right)$
$\displaystyle 4a + 12 = - 2a^2 + 14a$
$\displaystyle 2a^2 - 10a + 12 = 0$
$\displaystyle 2\left( {a^2 - 5a + 6} \right) = 0$
$\displaystyle 2\left( {a - 3} \right)\left( {a - 2} \right) = 0$
$\displaystyle a - 3 = 0 \Rightarrow a = 3$
or
$\displaystyle a - 2 = 0 \Rightarrow a = 2$
Just like the last problem, we have equal ratios, so we cross multiply:
$\displaystyle 4(a + 3) = -2a(a - 7)$
Evaluating, we get:
$\displaystyle 4a + 12 = -2a^2 + 14a$
Bring the right side over to the left side, we get:
$\displaystyle 2a^2 - 10a + 12 = 0$
That's a quadratic equation. If you need to solve that, go here and enter your coefficients:
Quadratic Equation
I get a = (2, 3).
Plug that back into your original proportion and see if it works.