Solve (-k/2)^2-7=3

$$( - \frac { k } { 2 } ) ^ { 2 } - 7 = 3$$

Solve for k

$$k=-2\sqrt{10}\approx -6.32455532$$

$$k=2\sqrt{10}\approx 6.32455532$$

Steps by Finding Square Root

Calculate -\frac{k}{2} to the power of 2 and get \left(\frac{k}{2}\right)^{2}.

$$\left(\frac{k}{2}\right)^{2}-7=3$$

To raise \frac{k}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{k^{2}}{2^{2}}-7=3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{2^{2}}{2^{2}}.

$$\frac{k^{2}}{2^{2}}-\frac{7\times 2^{2}}{2^{2}}=3$$

Since \frac{k^{2}}{2^{2}} and \frac{7\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{k^{2}-7\times 2^{2}}{2^{2}}=3$$

Do the multiplications in k^{2}-7\times 2^{2}.

$$\frac{k^{2}-28}{2^{2}}=3$$

Calculate 2 to the power of 2 and get 4.

$$\frac{k^{2}-28}{4}=3$$

Divide each term of k^{2}-28 by 4 to get \frac{1}{4}k^{2}-7.

$$\frac{1}{4}k^{2}-7=3$$

Add 7 to both sides.

$$\frac{1}{4}k^{2}=3+7$$

Add 3 and 7 to get 10.

$$\frac{1}{4}k^{2}=10$$

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$k^{2}=10\times 4$$

Multiply 10 and 4 to get 40.

$$k^{2}=40$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Calculate -\frac{k}{2} to the power of 2 and get \left(\frac{k}{2}\right)^{2}.

$$\left(\frac{k}{2}\right)^{2}-7=3$$

To raise \frac{k}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{k^{2}}{2^{2}}-7=3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{2^{2}}{2^{2}}.

$$\frac{k^{2}}{2^{2}}-\frac{7\times 2^{2}}{2^{2}}=3$$

Since \frac{k^{2}}{2^{2}} and \frac{7\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{k^{2}-7\times 2^{2}}{2^{2}}=3$$

Do the multiplications in k^{2}-7\times 2^{2}.

$$\frac{k^{2}-28}{2^{2}}=3$$

Subtract 3 from both sides.

$$\frac{k^{2}-28}{2^{2}}-3=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{k^{2}-28}{4}-3=0$$

Divide each term of k^{2}-28 by 4 to get \frac{1}{4}k^{2}-7.

$$\frac{1}{4}k^{2}-7-3=0$$

Subtract 3 from -7 to get -10.

$$\frac{1}{4}k^{2}-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, 0 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\times \frac{1}{4}\left(-10\right)}}{2\times \frac{1}{4}}$$

Square 0.

$$k=\frac{0±\sqrt{-4\times \frac{1}{4}\left(-10\right)}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$k=\frac{0±\sqrt{-\left(-10\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -10.

$$k=\frac{0±\sqrt{10}}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$k=\frac{0±\sqrt{10}}{\frac{1}{2}}$$

Now solve the equation k=\frac{0±\sqrt{10}}{\frac{1}{2}} when ± is plus.

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

Now solve the equation k=\frac{0±\sqrt{10}}{\frac{1}{2}} when ± is minus.

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

The equation is now solved.

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