Solve the equation x^2 - 6x - 5 = 0.

Step-by-step solution

1. 1

   Identify the coefficients and test for factoring. Here $a=1$, $b=-6$, $c=-5$. Integer factoring would need two numbers with product $-5$ and sum $-6$; the only integer pairs for $-5$ are $(1,-5)$ and $(5,-1)$, giving sums $-4$ and $4$ — neither is $-6$. So the roots are not integers and the formula is required.

2. 2

   Compute the discriminant.

   $\Delta=b^{2}-4ac=(-6)^{2}-4(1)(-5)=36+20=56$

   The $-4ac$ term becomes $+20$ because $c$ is negative — sign slips here are the commonest cause of a wrong answer. Since $\Delta=56>0$ but is not a perfect square, expect two irrational roots.

3. 3

   Substitute into the quadratic formula.

   $x=\frac{-(-6)\pm\sqrt{56}}{2(1)}=\frac{6\pm\sqrt{56}}{2}$

4. 4

   Simplify the surd, then cancel. Extract the largest square factor: $56=4\cdot14$, so $\sqrt{56}=2\sqrt{14}$. Now the common factor $2$ is visible:

   $x=\frac{6\pm2\sqrt{14}}{2}=\frac{2\left(3\pm\sqrt{14}\right)}{2}=3\pm\sqrt{14}$

   The $2$ must cancel from both numerator terms; cancelling it only against the $6$ would give the wrong $3\pm2\sqrt{14}$.

5. 5

   Verify with Vieta and numerically. The roots should sum to $-b/a=6$: $\left(3+\sqrt{14}\right)+\left(3-\sqrt{14}\right)=6$ ✓. They should multiply to $c/a=-5$: $3^{2}-\left(\sqrt{14}\right)^{2}=9-14=-5$ ✓. Numerically $x\approx6.7417$ and $x\approx-0.7417$, both giving residuals below $10^{-11}$ ✓.