Solve (k+2)^2-3(k+2)=5k+4

$$( k + 2 ) ^ { 2 } - 3 ( k + 2 ) = 5 k + 4$$

Solve for k

$$k=\sqrt{10}+2\approx 5.16227766$$

$$k=2-\sqrt{10}\approx -1.16227766$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k+2\right)^{2}.

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

Use the distributive property to multiply -3 by k+2.

$$k^{2}+4k+4-3k-6=5k+4$$

Combine 4k and -3k to get k.

$$k^{2}+k+4-6=5k+4$$

Subtract 6 from 4 to get -2.

$$k^{2}+k-2=5k+4$$

Subtract 5k from both sides.

$$k^{2}+k-2-5k=4$$

Combine k and -5k to get -4k.

$$k^{2}-4k-2=4$$

Subtract 4 from both sides.

$$k^{2}-4k-2-4=0$$

Subtract 4 from -2 to get -6.

$$k^{2}-4k-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$k=\frac{-\left(-4\right)±\sqrt{16-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$k=\frac{-\left(-4\right)±\sqrt{16+24}}{2}$$

Add 16 to 24.

$$k=\frac{-\left(-4\right)±\sqrt{40}}{2}$$

Take the square root of 40.

$$k=\frac{-\left(-4\right)±2\sqrt{10}}{2}$$

The opposite of -4 is 4.

$$k=\frac{4±2\sqrt{10}}{2}$$

Now solve the equation k=\frac{4±2\sqrt{10}}{2} when ± is plus. Add 4 to 2\sqrt{10}.

$$k=\frac{2\sqrt{10}+4}{2}$$

Divide 4+2\sqrt{10} by 2.

$$k=\sqrt{10}+2$$

Now solve the equation k=\frac{4±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from 4.

$$k=\frac{4-2\sqrt{10}}{2}$$

Divide 4-2\sqrt{10} by 2.

$$k=2-\sqrt{10}$$

The equation is now solved.

$$k=\sqrt{10}+2 k=2-\sqrt{10}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(k+2\right)^{2}.

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

Use the distributive property to multiply -3 by k+2.

$$k^{2}+4k+4-3k-6=5k+4$$

Combine 4k and -3k to get k.

$$k^{2}+k+4-6=5k+4$$

Subtract 6 from 4 to get -2.

$$k^{2}+k-2=5k+4$$

Subtract 5k from both sides.

$$k^{2}+k-2-5k=4$$

Combine k and -5k to get -4k.

$$k^{2}-4k-2=4$$

Add 2 to both sides.

$$k^{2}-4k=4+2$$

Add 4 and 2 to get 6.

$$k^{2}-4k=6$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-4k+\left(-2\right)^{2}=6+\left(-2\right)^{2}$$

Square -2.

$$k^{2}-4k+4=6+4$$

Add 6 to 4.

$$k^{2}-4k+4=10$$

Factor k^{2}-4k+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-2\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-2\right)^{2}}=\sqrt{10}$$

Simplify.

$$k-2=\sqrt{10} k-2=-\sqrt{10}$$

Add 2 to both sides of the equation.

$$k=\sqrt{10}+2 k=2-\sqrt{10}$$