Solve -(k-3)*(3k-9/4)=27

$$-(k-3) \times ( \frac{ 3k-9 }{ 4 } )=27$$

Solve for k

$$k=3+6i$$

$$k=3-6i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$\left(-\left(k-3\right)\right)\left(3k-9\right)=108$$

To find the opposite of k-3, find the opposite of each term.

$$\left(-k-\left(-3\right)\right)\left(3k-9\right)=108$$

The opposite of -3 is 3.

$$\left(-k+3\right)\left(3k-9\right)=108$$

Apply the distributive property by multiplying each term of -k+3 by each term of 3k-9.

$$-3k^{2}+9k+9k-27=108$$

Combine 9k and 9k to get 18k.

$$-3k^{2}+18k-27=108$$

Subtract 108 from both sides.

$$-3k^{2}+18k-27-108=0$$

Subtract 108 from -27 to get -135.

$$-3k^{2}+18k-135=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 18 for b, and -135 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-18±\sqrt{18^{2}-4\left(-3\right)\left(-135\right)}}{2\left(-3\right)}$$

Square 18.

$$k=\frac{-18±\sqrt{324-4\left(-3\right)\left(-135\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$k=\frac{-18±\sqrt{324+12\left(-135\right)}}{2\left(-3\right)}$$

Multiply 12 times -135.

$$k=\frac{-18±\sqrt{324-1620}}{2\left(-3\right)}$$

Add 324 to -1620.

$$k=\frac{-18±\sqrt{-1296}}{2\left(-3\right)}$$

Take the square root of -1296.

$$k=\frac{-18±36i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$k=\frac{-18±36i}{-6}$$

Now solve the equation k=\frac{-18±36i}{-6} when ± is plus. Add -18 to 36i.

$$k=\frac{-18+36i}{-6}$$

Divide -18+36i by -6.

$$k=3-6i$$

Now solve the equation k=\frac{-18±36i}{-6} when ± is minus. Subtract 36i from -18.

$$k=\frac{-18-36i}{-6}$$

Divide -18-36i by -6.

$$k=3+6i$$

The equation is now solved.

$$k=3-6i k=3+6i$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$\left(-\left(k-3\right)\right)\left(3k-9\right)=108$$

To find the opposite of k-3, find the opposite of each term.

$$\left(-k-\left(-3\right)\right)\left(3k-9\right)=108$$

The opposite of -3 is 3.

$$\left(-k+3\right)\left(3k-9\right)=108$$

Apply the distributive property by multiplying each term of -k+3 by each term of 3k-9.

$$-3k^{2}+9k+9k-27=108$$

Combine 9k and 9k to get 18k.

$$-3k^{2}+18k-27=108$$

Add 27 to both sides.

$$-3k^{2}+18k=108+27$$

Add 108 and 27 to get 135.

$$-3k^{2}+18k=135$$

Divide both sides by -3.

$$\frac{-3k^{2}+18k}{-3}=\frac{135}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$k^{2}+\frac{18}{-3}k=\frac{135}{-3}$$

Divide 18 by -3.

$$k^{2}-6k=\frac{135}{-3}$$

Divide 135 by -3.

$$k^{2}-6k=-45$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-6k+\left(-3\right)^{2}=-45+\left(-3\right)^{2}$$

Square -3.

$$k^{2}-6k+9=-45+9$$

Add -45 to 9.

$$k^{2}-6k+9=-36$$

Factor k^{2}-6k+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-3\right)^{2}=-36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-3\right)^{2}}=\sqrt{-36}$$

Simplify.

$$k-3=6i k-3=-6i$$

Add 3 to both sides of the equation.

$$k=3+6i k=3-6i$$