Find the real solutions of the equation 7(s + 9)^2 - 20(s + 9) = 3.

Step-by-step solution

1. 1

   Spot the repeated block. The expression $s+9$ appears squared and to the first power, and nowhere else. That is the signal for a substitution: the equation is quadratic in the block, not in $s$ directly.

2. 2

   Let u = s + 9. The equation becomes $7u^2-20u=3$, and moving the $3$ across gives the standard form $7u^2-20u-3=0$. Expanding the square first would work too but produces larger numbers for no gain.

3. 3

   Factor the quadratic in u. Looking for two numbers multiplying to $7\cdot(-3)=-21$ and adding to $-20$ gives $-21$ and $1$, so $7u^2-21u+u-3=7u(u-3)+(u-3)=(7u+1)(u-3)=0$. Hence $u=3$ or $u=-\frac17$.

4. 4

   Undo the substitution. Since $u=s+9$, we get $s=u-9$. From $u=3$: $s=-6$. From $u=-\frac17$: $s=-\frac17-9=-\frac{64}{7}$.

5. 5

   Check both roots in the original equation. For $s=-6$: $s+9=3$, so $7(9)-20(3)=63-60=3$. For $s=-\frac{64}{7}$: $s+9=-\frac17$, so $7\cdot\frac{1}{49}+\frac{20}{7}=\frac17+\frac{20}{7}=3$. Both give exactly $3$.