Solve x2 + 3x + 6 = 1.

Step-by-step solution

1. 1

   Move the constant across to get standard form. Subtract $1$ from both sides:

   $x^2+3x+5=0$

   The constant becomes $5$, not $7$ — and skipping this step is the most common error, since the quadratic formula needs $c$ from the zero form.

2. 2

   Compute the discriminant. With $a=1$, $b=3$, $c=5$,

   $\Delta=3^2-4(1)(5)=9-20=-11$

   A negative discriminant means the parabola never reaches the $x$-axis: no real solutions. (Indeed the minimum of $x^2+3x+6$ is $6-\frac94=3.75$, well above $1$.)

3. 3

   Apply the quadratic formula anyway, over the complex numbers.

   $x=\frac{-3\pm\sqrt{-11}}{2}$

4. 4

   Rewrite the negative root using $i$. Since $\sqrt{-11}=i\sqrt{11}$,

   $x=\frac{-3\pm i\sqrt{11}}{2}$

   The two solutions are complex conjugates, which is always the case for a real quadratic with $\Delta<0$.

5. 5

   Verify with sum and product. The roots should sum to $-b/a=-3$: indeed $\frac{-3+i\sqrt{11}}{2}+\frac{-3-i\sqrt{11}}{2}=-3$ ✓. Their product should be $c/a=5$:

   $\frac{(-3)^2-\left(i\sqrt{11}\right)^2}{4}=\frac{9+11}{4}=5$ ✓