Solve 11^2*k^2-5k-6=0

$$11 ^ { 2 } \cdot k ^ { 2 } - 5 k - 6 = 0$$

Solve for k

$$k=\frac{\sqrt{2929}+5}{242}\approx 0.244298498$$

$$k=\frac{5-\sqrt{2929}}{242}\approx -0.202976184$$

Steps Using the Quadratic Formula

Calculate 11 to the power of 2 and get 121.

$$121k^{2}-5k-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 121 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 121\left(-6\right)}}{2\times 121}$$

Square -5.

$$k=\frac{-\left(-5\right)±\sqrt{25-4\times 121\left(-6\right)}}{2\times 121}$$

Multiply -4 times 121.

$$k=\frac{-\left(-5\right)±\sqrt{25-484\left(-6\right)}}{2\times 121}$$

Multiply -484 times -6.

$$k=\frac{-\left(-5\right)±\sqrt{25+2904}}{2\times 121}$$

Add 25 to 2904.

$$k=\frac{-\left(-5\right)±\sqrt{2929}}{2\times 121}$$

The opposite of -5 is 5.

$$k=\frac{5±\sqrt{2929}}{2\times 121}$$

Multiply 2 times 121.

$$k=\frac{5±\sqrt{2929}}{242}$$

Now solve the equation k=\frac{5±\sqrt{2929}}{242} when ± is plus. Add 5 to \sqrt{2929}.

$$k=\frac{\sqrt{2929}+5}{242}$$

Now solve the equation k=\frac{5±\sqrt{2929}}{242} when ± is minus. Subtract \sqrt{2929} from 5.

$$k=\frac{5-\sqrt{2929}}{242}$$

The equation is now solved.

$$k=\frac{\sqrt{2929}+5}{242} k=\frac{5-\sqrt{2929}}{242}$$

Steps for Completing the Square

Calculate 11 to the power of 2 and get 121.

$$121k^{2}-5k-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$121k^{2}-5k=6$$

Divide both sides by 121.

$$\frac{121k^{2}-5k}{121}=\frac{6}{121}$$

Dividing by 121 undoes the multiplication by 121.

$$k^{2}-\frac{5}{121}k=\frac{6}{121}$$

Divide -\frac{5}{121}, the coefficient of the x term, by 2 to get -\frac{5}{242}. Then add the square of -\frac{5}{242} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{5}{121}k+\left(-\frac{5}{242}\right)^{2}=\frac{6}{121}+\left(-\frac{5}{242}\right)^{2}$$

Square -\frac{5}{242} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{5}{121}k+\frac{25}{58564}=\frac{6}{121}+\frac{25}{58564}$$

Add \frac{6}{121} to \frac{25}{58564} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{5}{121}k+\frac{25}{58564}=\frac{2929}{58564}$$

Factor k^{2}-\frac{5}{121}k+\frac{25}{58564}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{5}{242}\right)^{2}=\frac{2929}{58564}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{5}{242}\right)^{2}}=\sqrt{\frac{2929}{58564}}$$

Simplify.

$$k-\frac{5}{242}=\frac{\sqrt{2929}}{242} k-\frac{5}{242}=-\frac{\sqrt{2929}}{242}$$

Add \frac{5}{242} to both sides of the equation.

$$k=\frac{\sqrt{2929}+5}{242} k=\frac{5-\sqrt{2929}}{242}$$