Solve (6/5)^2=2^2+C^2-2*2*C*4/5

$$( \frac { 6 } { 5 } ) ^ { 2 } = 2 ^ { 2 } + C ^ { 2 } - 2 \times 2 \times C \times \frac { 4 } { 5 }$$

Solve for C

$$C = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

Steps Using the Quadratic Formula

Calculate \frac{6}{5} to the power of 2 and get \frac{36}{25}.

$$\frac{36}{25}=2^{2}+C^{2}-2\times 2C\times \frac{4}{5}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{36}{25}=4+C^{2}-2\times 2C\times \frac{4}{5}$$

Multiply 2 and 2 to get 4.

$$\frac{36}{25}=4+C^{2}-4C\times \frac{4}{5}$$

Multiply 4 and \frac{4}{5} to get \frac{16}{5}.

$$\frac{36}{25}=4+C^{2}-\frac{16}{5}C$$

Swap sides so that all variable terms are on the left hand side.

$$4+C^{2}-\frac{16}{5}C=\frac{36}{25}$$

Subtract \frac{36}{25} from both sides.

$$4+C^{2}-\frac{16}{5}C-\frac{36}{25}=0$$

Subtract \frac{36}{25} from 4 to get \frac{64}{25}.

$$\frac{64}{25}+C^{2}-\frac{16}{5}C=0$$

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.

$$C^{2}-\frac{16}{5}C+\frac{64}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{16}{5} for b, and \frac{64}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$C=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\left(-\frac{16}{5}\right)^{2}-4\times \frac{64}{25}}}{2}$$

Square -\frac{16}{5} by squaring both the numerator and the denominator of the fraction.

$$C=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\frac{256}{25}-4\times \frac{64}{25}}}{2}$$

Multiply -4 times \frac{64}{25}.

$$C=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\frac{256-256}{25}}}{2}$$

Add \frac{256}{25} to -\frac{256}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$C=\frac{-\left(-\frac{16}{5}\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$C=-\frac{-\frac{16}{5}}{2}$$

The opposite of -\frac{16}{5} is \frac{16}{5}.

$$C=\frac{\frac{16}{5}}{2}$$

Divide \frac{16}{5} by 2.

$$C=\frac{8}{5}$$

Steps for Completing the Square

Calculate \frac{6}{5} to the power of 2 and get \frac{36}{25}.

$$\frac{36}{25}=2^{2}+C^{2}-2\times 2C\times \frac{4}{5}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{36}{25}=4+C^{2}-2\times 2C\times \frac{4}{5}$$

Multiply 2 and 2 to get 4.

$$\frac{36}{25}=4+C^{2}-4C\times \frac{4}{5}$$

Multiply 4 and \frac{4}{5} to get \frac{16}{5}.

$$\frac{36}{25}=4+C^{2}-\frac{16}{5}C$$

Swap sides so that all variable terms are on the left hand side.

$$4+C^{2}-\frac{16}{5}C=\frac{36}{25}$$

Subtract 4 from both sides.

$$C^{2}-\frac{16}{5}C=\frac{36}{25}-4$$

Subtract 4 from \frac{36}{25} to get -\frac{64}{25}.

$$C^{2}-\frac{16}{5}C=-\frac{64}{25}$$

Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$C^{2}-\frac{16}{5}C+\left(-\frac{8}{5}\right)^{2}=-\frac{64}{25}+\left(-\frac{8}{5}\right)^{2}$$

Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$C^{2}-\frac{16}{5}C+\frac{64}{25}=\frac{-64+64}{25}$$

Add -\frac{64}{25} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$C^{2}-\frac{16}{5}C+\frac{64}{25}=0$$

Factor C^{2}-\frac{16}{5}C+\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(C-\frac{8}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(C-\frac{8}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$C-\frac{8}{5}=0 C-\frac{8}{5}=0$$

Add \frac{8}{5} to both sides of the equation.

$$C=\frac{8}{5} C=\frac{8}{5}$$

The equation is now solved. Solutions are the same.

$$C=\frac{8}{5}$$