Solve the equation x^2 = 2x + 17.

Step-by-step solution

1. 1

   Rearrange to standard form. Subtract $2x$ and $17$ from both sides so that one side is zero:

   $x^{2}-2x-17=0$

   so $a=1$, $b=-2$, $c=-17$. Both moved terms change sign.

2. 2

   Compute the discriminant.

   $\Delta=(-2)^{2}-4(1)(-17)=4+68=72$

   Since $\Delta=72>0$ there are two distinct real roots. It is not a perfect square, so factoring over the integers is impossible — no pair of integers multiplies to $-17$ and adds to $-2$, as $17$ is prime.

3. 3

   Apply the quadratic formula.

   $x=\frac{-(-2)\pm\sqrt{72}}{2(1)}=\frac{2\pm\sqrt{72}}{2}$

4. 4

   Simplify the radical before cancelling. Extract the largest square factor: $72=36\cdot2$, so $\sqrt{72}=6\sqrt{2}$. Then

   $x=\frac{2\pm6\sqrt{2}}{2}=\frac{2\left(1\pm3\sqrt{2}\right)}{2}=1\pm3\sqrt{2}$

   Leaving $\sqrt{72}$ unsimplified hides the common factor $2$ and blocks the clean cancellation.

5. 5

   Verify both roots. Vieta: the sum should be $-b/a=2$, and $\left(1+3\sqrt2\right)+\left(1-3\sqrt2\right)=2$ ✓. The product should be $c/a=-17$, and $1^{2}-\left(3\sqrt2\right)^{2}=1-18=-17$ ✓. Numerically $x\approx5.2426$ and $x\approx-3.2426$, both giving residuals below $10^{-11}$ in $x^{2}-2x-17$ ✓.