Solve (k-2)^2-4*(-1)*(2k-9)=0

$$( k - 2 ) ^ { 2 } - 4 \times ( - 1 ) \times ( 2 k - 9 ) = 0$$

Solve for k

$$k=-8$$

$$k=4$$

Steps Using Factoring

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

$$k^{2}-4k+4-4\left(-1\right)\left(2k-9\right)=0$$

Multiply 4 and -1 to get -4.

$$k^{2}-4k+4-\left(-4\left(2k-9\right)\right)=0$$

The opposite of -4\left(2k-9\right) is 4\left(2k-9\right).

$$k^{2}-4k+4+4\left(2k-9\right)=0$$

Use the distributive property to multiply 4 by 2k-9.

$$k^{2}-4k+4+8k-36=0$$

Combine -4k and 8k to get 4k.

$$k^{2}+4k+4-36=0$$

Subtract 36 from 4 to get -32.

$$k^{2}+4k-32=0$$

To solve the equation, factor k^{2}+4k-32 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-32$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$a=-4 b=8$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k-4\right)\left(k+8\right)$$

To find equation solutions, solve k-4=0 and k+8=0.

$$k=4 k=-8$$

Steps Using Factoring By Grouping

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

$$k^{2}-4k+4-4\left(-1\right)\left(2k-9\right)=0$$

Multiply 4 and -1 to get -4.

$$k^{2}-4k+4-\left(-4\left(2k-9\right)\right)=0$$

The opposite of -4\left(2k-9\right) is 4\left(2k-9\right).

$$k^{2}-4k+4+4\left(2k-9\right)=0$$

Use the distributive property to multiply 4 by 2k-9.

$$k^{2}-4k+4+8k-36=0$$

Combine -4k and 8k to get 4k.

$$k^{2}+4k+4-36=0$$

Subtract 36 from 4 to get -32.

$$k^{2}+4k-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk-32. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-32\right)=-32$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$a=-4 b=8$$

Rewrite k^{2}+4k-32 as \left(k^{2}-4k\right)+\left(8k-32\right).

$$\left(k^{2}-4k\right)+\left(8k-32\right)$$

Factor out k in the first and 8 in the second group.

$$k\left(k-4\right)+8\left(k-4\right)$$

Factor out common term k-4 by using distributive property.

$$\left(k-4\right)\left(k+8\right)$$

To find equation solutions, solve k-4=0 and k+8=0.

$$k=4 k=-8$$

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}.

$$k^{2}-4k+4-4\left(-1\right)\left(2k-9\right)=0$$

Multiply 4 and -1 to get -4.

$$k^{2}-4k+4-\left(-4\left(2k-9\right)\right)=0$$

The opposite of -4\left(2k-9\right) is 4\left(2k-9\right).

$$k^{2}-4k+4+4\left(2k-9\right)=0$$

Use the distributive property to multiply 4 by 2k-9.

$$k^{2}-4k+4+8k-36=0$$

Combine -4k and 8k to get 4k.

$$k^{2}+4k+4-36=0$$

Subtract 36 from 4 to get -32.

$$k^{2}+4k-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-4±\sqrt{4^{2}-4\left(-32\right)}}{2}$$

Square 4.

$$k=\frac{-4±\sqrt{16-4\left(-32\right)}}{2}$$

Multiply -4 times -32.

$$k=\frac{-4±\sqrt{16+128}}{2}$$

Add 16 to 128.

$$k=\frac{-4±\sqrt{144}}{2}$$

Take the square root of 144.

$$k=\frac{-4±12}{2}$$

Now solve the equation k=\frac{-4±12}{2} when ± is plus. Add -4 to 12.

$$k=\frac{8}{2}$$

Divide 8 by 2.

$$k=4$$

Now solve the equation k=\frac{-4±12}{2} when ± is minus. Subtract 12 from -4.

$$k=-\frac{16}{2}$$

Divide -16 by 2.

$$k=-8$$

The equation is now solved.

$$k=4 k=-8$$

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}.

$$k^{2}-4k+4-4\left(-1\right)\left(2k-9\right)=0$$

Multiply 4 and -1 to get -4.

$$k^{2}-4k+4-\left(-4\left(2k-9\right)\right)=0$$

The opposite of -4\left(2k-9\right) is 4\left(2k-9\right).

$$k^{2}-4k+4+4\left(2k-9\right)=0$$

Use the distributive property to multiply 4 by 2k-9.

$$k^{2}-4k+4+8k-36=0$$

Combine -4k and 8k to get 4k.

$$k^{2}+4k+4-36=0$$

Subtract 36 from 4 to get -32.

$$k^{2}+4k-32=0$$

Add 32 to both sides. Anything plus zero gives itself.

$$k^{2}+4k=32$$

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.

$$k^{2}+4k+2^{2}=32+2^{2}$$

Square 2.

$$k^{2}+4k+4=32+4$$

Add 32 to 4.

$$k^{2}+4k+4=36$$

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}.

$$\left(k+2\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+2\right)^{2}}=\sqrt{36}$$

Simplify.

$$k+2=6 k+2=-6$$

Subtract 2 from both sides of the equation.

$$k=4 k=-8$$