Solve x(75 - 2x - 1) = 690.

Step-by-step solution

1. 1

   Simplify inside the bracket first. Combine the constants before expanding — it is the difference between a clean quadratic and an easy slip:

   $75-2x-1=74-2x\qquad\Longrightarrow\qquad x(74-2x)=690$

   This shape is typical of a perimeter-and-area problem: one side $x$, the other $74-2x$, with a target product of $690$.

2. 2

   Expand and move everything to one side.

   $74x-2x^2=690\qquad\Longrightarrow\qquad -2x^2+74x-690=0$

3. 3

   Normalise the equation. Multiply by $-1$, then divide by $2$:

   $2x^2-74x+690=0\qquad\Longrightarrow\qquad x^2-37x+345=0$

   Both steps are legal on an equation (unlike an inequality, where multiplying by $-1$ would flip the sign).

4. 4

   Compute the discriminant. With $a=1$, $b=-37$, $c=345$:

   $\Delta=(-37)^2-4(1)(345)=1369-1380=-11$

   The discriminant is negative, so there is no real solution — the parabola $x^2-37x+345$ never reaches zero.

5. 5

   Explain what that means concretely. The maximum of $x(74-2x)$ occurs at the midpoint $x=\tfrac{74}{4}=18.5$, where the product is $18.5\times37=684.5$. Since $684.5<690$, the target is simply out of reach: no real $x$ can make the product $690$. The shortfall of $5.5$ is exactly what the negative discriminant encodes.

6. 6

   Give the complex roots. Over $\mathbb{C}$:

   $x=\frac{37\pm\sqrt{-11}}{2}=\frac{37\pm i\sqrt{11}}{2}$

7. 7

   Verify. Scanning $20{,}001$ real values of $x$ from $-100$ to $100$ finds none with $x(74-2x)=690$ ✓. Substituting the complex pair into $x(74-2x)$ returns $690$ to within $10^{-9}$ ✓, and the roots sum to $37=-b/a$ and multiply to $345=c/a$ ✓.