Solve 3(k-2)-6=4k(3k-1)

$$3 ( k - 2 ) - 6 = 4 k ( 3 k - 1 )$$

Solve for k

$$k=\frac{-\sqrt{527}i+7}{24}\approx 0.291666667-0.956520024i$$

$$k=\frac{7+\sqrt{527}i}{24}\approx 0.291666667+0.956520024i$$

Steps Using the Quadratic Formula

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

$$3k-6-6=4k\left(3k-1\right)$$

Subtract 6 from -6 to get -12.

$$3k-12=4k\left(3k-1\right)$$

Use the distributive property to multiply 4k by 3k-1.

$$3k-12=12k^{2}-4k$$

Subtract 12k^{2} from both sides.

$$3k-12-12k^{2}=-4k$$

Add 4k to both sides.

$$3k-12-12k^{2}+4k=0$$

Combine 3k and 4k to get 7k.

$$7k-12-12k^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-12k^{2}+7k-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 7 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-7±\sqrt{7^{2}-4\left(-12\right)\left(-12\right)}}{2\left(-12\right)}$$

Square 7.

$$k=\frac{-7±\sqrt{49-4\left(-12\right)\left(-12\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$k=\frac{-7±\sqrt{49+48\left(-12\right)}}{2\left(-12\right)}$$

Multiply 48 times -12.

$$k=\frac{-7±\sqrt{49-576}}{2\left(-12\right)}$$

Add 49 to -576.

$$k=\frac{-7±\sqrt{-527}}{2\left(-12\right)}$$

Take the square root of -527.

$$k=\frac{-7±\sqrt{527}i}{2\left(-12\right)}$$

Multiply 2 times -12.

$$k=\frac{-7±\sqrt{527}i}{-24}$$

Now solve the equation k=\frac{-7±\sqrt{527}i}{-24} when ± is plus. Add -7 to i\sqrt{527}.

$$k=\frac{-7+\sqrt{527}i}{-24}$$

Divide -7+i\sqrt{527} by -24.

$$k=\frac{-\sqrt{527}i+7}{24}$$

Now solve the equation k=\frac{-7±\sqrt{527}i}{-24} when ± is minus. Subtract i\sqrt{527} from -7.

$$k=\frac{-\sqrt{527}i-7}{-24}$$

Divide -7-i\sqrt{527} by -24.

$$k=\frac{7+\sqrt{527}i}{24}$$

The equation is now solved.

$$k=\frac{-\sqrt{527}i+7}{24} k=\frac{7+\sqrt{527}i}{24}$$

Steps for Completing the Square

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

$$3k-6-6=4k\left(3k-1\right)$$

Subtract 6 from -6 to get -12.

$$3k-12=4k\left(3k-1\right)$$

Use the distributive property to multiply 4k by 3k-1.

$$3k-12=12k^{2}-4k$$

Subtract 12k^{2} from both sides.

$$3k-12-12k^{2}=-4k$$

Add 4k to both sides.

$$3k-12-12k^{2}+4k=0$$

Combine 3k and 4k to get 7k.

$$7k-12-12k^{2}=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$7k-12k^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-12k^{2}+7k=12$$

Divide both sides by -12.

$$\frac{-12k^{2}+7k}{-12}=\frac{12}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$k^{2}+\frac{7}{-12}k=\frac{12}{-12}$$

Divide 7 by -12.

$$k^{2}-\frac{7}{12}k=\frac{12}{-12}$$

Divide 12 by -12.

$$k^{2}-\frac{7}{12}k=-1$$

Divide -\frac{7}{12}, the coefficient of the x term, by 2 to get -\frac{7}{24}. Then add the square of -\frac{7}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{7}{12}k+\left(-\frac{7}{24}\right)^{2}=-1+\left(-\frac{7}{24}\right)^{2}$$

Square -\frac{7}{24} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{7}{12}k+\frac{49}{576}=-1+\frac{49}{576}$$

Add -1 to \frac{49}{576}.

$$k^{2}-\frac{7}{12}k+\frac{49}{576}=-\frac{527}{576}$$

Factor k^{2}-\frac{7}{12}k+\frac{49}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{7}{24}\right)^{2}=-\frac{527}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{7}{24}\right)^{2}}=\sqrt{-\frac{527}{576}}$$

Simplify.

$$k-\frac{7}{24}=\frac{\sqrt{527}i}{24} k-\frac{7}{24}=-\frac{\sqrt{527}i}{24}$$

Add \frac{7}{24} to both sides of the equation.

$$k=\frac{7+\sqrt{527}i}{24} k=\frac{-\sqrt{527}i+7}{24}$$