Solve 2k^2-4(4-3k)=0

$$2 k ^ { 2 } - 4 ( 4 - 3 k ) = 0$$

Solve for k (complex solution)

$$k=\sqrt{17}-3\approx 1.123105626$$

$$k=-\left(\sqrt{17}+3\right)\approx -7.123105626$$

Steps Using the Quadratic Formula

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

$$2k^{2}-16+12k=0$$

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.

$$2k^{2}+12k-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-12±\sqrt{12^{2}-4\times 2\left(-16\right)}}{2\times 2}$$

Square 12.

$$k=\frac{-12±\sqrt{144-4\times 2\left(-16\right)}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-12±\sqrt{144-8\left(-16\right)}}{2\times 2}$$

Multiply -8 times -16.

$$k=\frac{-12±\sqrt{144+128}}{2\times 2}$$

Add 144 to 128.

$$k=\frac{-12±\sqrt{272}}{2\times 2}$$

Take the square root of 272.

$$k=\frac{-12±4\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{-12±4\sqrt{17}}{4}$$

Now solve the equation k=\frac{-12±4\sqrt{17}}{4} when ± is plus. Add -12 to 4\sqrt{17}.

$$k=\frac{4\sqrt{17}-12}{4}$$

Divide -12+4\sqrt{17} by 4.

$$k=\sqrt{17}-3$$

Now solve the equation k=\frac{-12±4\sqrt{17}}{4} when ± is minus. Subtract 4\sqrt{17} from -12.

$$k=\frac{-4\sqrt{17}-12}{4}$$

Divide -12-4\sqrt{17} by 4.

$$k=-\sqrt{17}-3$$

The equation is now solved.

$$k=\sqrt{17}-3 k=-\sqrt{17}-3$$

Steps for Completing the Square

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

$$2k^{2}-16+12k=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$2k^{2}+12k=16$$

Divide both sides by 2.

$$\frac{2k^{2}+12k}{2}=\frac{16}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\frac{12}{2}k=\frac{16}{2}$$

Divide 12 by 2.

$$k^{2}+6k=\frac{16}{2}$$

Divide 16 by 2.

$$k^{2}+6k=8$$

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.

$$k^{2}+6k+3^{2}=8+3^{2}$$

Square 3.

$$k^{2}+6k+9=8+9$$

Add 8 to 9.

$$k^{2}+6k+9=17$$

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}.

$$\left(k+3\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+3\right)^{2}}=\sqrt{17}$$

Simplify.

$$k+3=\sqrt{17} k+3=-\sqrt{17}$$

Subtract 3 from both sides of the equation.

$$k=\sqrt{17}-3 k=-\sqrt{17}-3$$

Solve for k

$$k=\sqrt{17}-3\approx 1.123105626$$

$$k=-\sqrt{17}-3\approx -7.123105626$$

Steps Using the Quadratic Formula

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

$$2k^{2}-16+12k=0$$

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.

$$2k^{2}+12k-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-12±\sqrt{12^{2}-4\times 2\left(-16\right)}}{2\times 2}$$

Square 12.

$$k=\frac{-12±\sqrt{144-4\times 2\left(-16\right)}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-12±\sqrt{144-8\left(-16\right)}}{2\times 2}$$

Multiply -8 times -16.

$$k=\frac{-12±\sqrt{144+128}}{2\times 2}$$

Add 144 to 128.

$$k=\frac{-12±\sqrt{272}}{2\times 2}$$

Take the square root of 272.

$$k=\frac{-12±4\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{-12±4\sqrt{17}}{4}$$

Now solve the equation k=\frac{-12±4\sqrt{17}}{4} when ± is plus. Add -12 to 4\sqrt{17}.

$$k=\frac{4\sqrt{17}-12}{4}$$

Divide -12+4\sqrt{17} by 4.

$$k=\sqrt{17}-3$$

Now solve the equation k=\frac{-12±4\sqrt{17}}{4} when ± is minus. Subtract 4\sqrt{17} from -12.

$$k=\frac{-4\sqrt{17}-12}{4}$$

Divide -12-4\sqrt{17} by 4.

$$k=-\sqrt{17}-3$$

The equation is now solved.

$$k=\sqrt{17}-3 k=-\sqrt{17}-3$$

Steps for Completing the Square

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

$$2k^{2}-16+12k=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$2k^{2}+12k=16$$

Divide both sides by 2.

$$\frac{2k^{2}+12k}{2}=\frac{16}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\frac{12}{2}k=\frac{16}{2}$$

Divide 12 by 2.

$$k^{2}+6k=\frac{16}{2}$$

Divide 16 by 2.

$$k^{2}+6k=8$$

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.

$$k^{2}+6k+3^{2}=8+3^{2}$$

Square 3.

$$k^{2}+6k+9=8+9$$

Add 8 to 9.

$$k^{2}+6k+9=17$$

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}.

$$\left(k+3\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+3\right)^{2}}=\sqrt{17}$$

Simplify.

$$k+3=\sqrt{17} k+3=-\sqrt{17}$$

Subtract 3 from both sides of the equation.

$$k=\sqrt{17}-3 k=-\sqrt{17}-3$$