Solve the quadratic equations using the complete the square method: 1) x2 + 3x + 1 = 0, 2) x2 − 7x − 2 = 0

Okay, ~DÅRK, let's solve these using the complete the square method again.

1) $x^2 + 3x + 1 = 0$

Step 1: Move the constant term to the right side.

$$x^2 + 3x = -1$$

Step 2: Complete the square on the left side by adding $(\frac{3}{2})^2 = \frac{9}{4}$ to both sides.

$$x^2 + 3x + \frac{9}{4} = -1 + \frac{9}{4}$$

Step 3: Factor the left side as a perfect square and simplify the right side.

$$\left(x + \frac{3}{2}\right)^2 = \frac{-4}{4} + \frac{9}{4}$$ $$\left(x + \frac{3}{2}\right)^2 = \frac{5}{4}$$

Step 4: Take the square root of both sides.

$$x + \frac{3}{2} = \pm\sqrt{\frac{5}{4}}$$ $$x + \frac{3}{2} = \pm\frac{\sqrt{5}}{2}$$

Step 5: Solve for $x$.

$$x = -\frac{3}{2} \pm \frac{\sqrt{5}}{2}$$ $$x = \frac{-3 \pm \sqrt{5}}{2}$$

The solutions are $\boxed{x = \frac{-3 \pm \sqrt{5}}{2}}$.

2) $x^2 - 7x - 2 = 0$

Step 1: Move the constant term to the right side.

$$x^2 - 7x = 2$$

Step 2: Complete the square on the left side by adding $(-\frac{7}{2})^2 = \frac{49}{4}$ to both sides.

$$x^2 - 7x + \frac{49}{4} = 2 + \frac{49}{4}$$

Step 3: Factor the left side as a perfect square and simplify the right side.

$$\left(x - \frac{7}{2}\right)^2 = \frac{8}{4} + \frac{49}{4}$$ $$\left(x - \frac{7}{2}\right)^2 = \frac{57}{4}$$

Step 4: Take the square root of both sides.

$$x - \frac{7}{2} = \pm\sqrt{\frac{57}{4}}$$ $$x - \frac{7}{2} = \pm\frac{\sqrt{57}}{2}$$

Step 5: Solve for $x$.

$$x = \frac{7}{2} \pm \frac{\sqrt{57}}{2}$$ $$x = \frac{7 \pm \sqrt{57}}{2}$$

The solutions are $\boxed{x = \frac{7 \pm \sqrt{57}}{2}}$.

3) $2m^2 + 10m - 5 = 0$

Step 1: Divide the entire equation by the coefficient of $m^2$, which is 2.

$$m^2 + 5m - \frac{5}{2} = 0$$

Step 2: Move the constant term to the right side.

$$m^2 + 5m = \frac{5}{2}$$

Step 3: Complete the square on the left side by adding $(\frac{5}{2})^2 = \frac{25}{4}$ to both sides.

$$m^2 + 5m + \frac{25}{4} = \frac{5}{2} + \frac{25}{4}$$

Step 4: Factor the left side as a perfect square and simplify the right side.

$$\left(m + \frac{5}{2}\right)^2 = \frac{10}{4} + \frac{25}{4}$$ $$\left(m + \frac{5}{2}\right)^2 = \frac{35}{4}$$

Step 5: Take the square root of both sides.

$$m + \frac{5}{2} = \pm\sqrt{\frac{35}{4}}$$ $$m + \frac{5}{2} = \pm\frac{\sqrt{35}}{2}$$

Step 6: Solve for $m$.

$$m = -\frac{5}{2} \pm \frac{\sqrt{35}}{2}$$ $$m = \frac{-5 \pm \sqrt{35}}{2}$$

The solutions are $\boxed{m = \frac{-5 \pm \sqrt{35}}{2}}$.

4) $5q^2 - q - 3 = 0$

Step 1: Divide the entire equation by the coefficient of $q^2$, which is 5.

$$q^2 - \frac{1}{5}q - \frac{3}{5} = 0$$

Step 2: Move the constant term to the right side.

$$q^2 - \frac{1}{5}q = \frac{3}{5}$$

Step 3: Complete the square on the left side by adding $(-\frac{1}{10})^2 = \frac{1}{100}$ to both sides.

$$q^2 - \frac{1}{5}q + \frac{1}{100} = \frac{3}{5} + \frac{1}{100}$$

Step 4: Factor the left side as a perfect square and simplify the right side.

$$\left(q - \frac{1}{10}\right)^2 = \frac{60}{100} + \frac{1}{100}$$ $$\left(q - \frac{1}{10}\right)^2 = \frac{61}{100}$$

Step 5: Take the square root of both sides.

$$q - \frac{1}{10} = \pm\sqrt{\frac{61}{100}}$$ $$q - \frac{1}{10} = \pm\frac{\sqrt{61}}{10}$$

Step 6: Solve for $q$.

$$q = \frac{1}{10} \pm \frac{\sqrt{61}}{10}$$ $$q = \frac{1 \pm \sqrt{61}}{10}$$

The solutions are $\boxed{q = \frac{1 \pm \sqrt{61}}{10}}$.

5) $10x^2 - 4x = 16$

Step 1: Rewrite the equation in standard form and divide by the coefficient of $x^2$, which is 10.

$$10x^2 - 4x - 16 = 0$$ $$$$