Solve (x-3)^2+(x/4+7/2)^2=85

$$( x - 3 ) ^ { 2 } + ( \frac { x } { 4 } + \frac { 7 } { 2 } ) ^ { 2 } = 85$$

Solve for x

$$x=-6$$

$$x=10$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+\left(\frac{x}{4}+\frac{7}{2}\right)^{2}=85$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{7}{2} times \frac{2}{2}.

$$x^{2}-6x+9+\left(\frac{x}{4}+\frac{7\times 2}{4}\right)^{2}=85$$

Since \frac{x}{4} and \frac{7\times 2}{4} have the same denominator, add them by adding their numerators.

$$x^{2}-6x+9+\left(\frac{x+7\times 2}{4}\right)^{2}=85$$

Do the multiplications in x+7\times 2.

$$x^{2}-6x+9+\left(\frac{x+14}{4}\right)^{2}=85$$

To raise \frac{x+14}{4} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-6x+9+\frac{\left(x+14\right)^{2}}{4^{2}}=85$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-6x+9 times \frac{4^{2}}{4^{2}}.

$$\frac{\left(x^{2}-6x+9\right)\times 4^{2}}{4^{2}}+\frac{\left(x+14\right)^{2}}{4^{2}}=85$$

Since \frac{\left(x^{2}-6x+9\right)\times 4^{2}}{4^{2}} and \frac{\left(x+14\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-6x+9\right)\times 4^{2}+\left(x+14\right)^{2}}{4^{2}}=85$$

Do the multiplications in \left(x^{2}-6x+9\right)\times 4^{2}+\left(x+14\right)^{2}.

$$\frac{16x^{2}-96x+144+x^{2}+28x+196}{4^{2}}=85$$

Combine like terms in 16x^{2}-96x+144+x^{2}+28x+196.

$$\frac{17x^{2}-68x+340}{4^{2}}=85$$

Calculate 4 to the power of 2 and get 16.

$$\frac{17x^{2}-68x+340}{16}=85$$

Divide each term of 17x^{2}-68x+340 by 16 to get \frac{17}{16}x^{2}-\frac{17}{4}x+\frac{85}{4}.

$$\frac{17}{16}x^{2}-\frac{17}{4}x+\frac{85}{4}=85$$

Subtract 85 from both sides.

$$\frac{17}{16}x^{2}-\frac{17}{4}x+\frac{85}{4}-85=0$$

Subtract 85 from \frac{85}{4} to get -\frac{255}{4}.

$$\frac{17}{16}x^{2}-\frac{17}{4}x-\frac{255}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{17}{16} for a, -\frac{17}{4} for b, and -\frac{255}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{17}{4}\right)±\sqrt{\left(-\frac{17}{4}\right)^{2}-4\times \frac{17}{16}\left(-\frac{255}{4}\right)}}{2\times \frac{17}{16}}$$

Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{17}{4}\right)±\sqrt{\frac{289}{16}-4\times \frac{17}{16}\left(-\frac{255}{4}\right)}}{2\times \frac{17}{16}}$$

Multiply -4 times \frac{17}{16}.

$$x=\frac{-\left(-\frac{17}{4}\right)±\sqrt{\frac{289}{16}-\frac{17}{4}\left(-\frac{255}{4}\right)}}{2\times \frac{17}{16}}$$

Multiply -\frac{17}{4} times -\frac{255}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{17}{4}\right)±\sqrt{\frac{289+4335}{16}}}{2\times \frac{17}{16}}$$

Add \frac{289}{16} to \frac{4335}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{17}{4}\right)±\sqrt{289}}{2\times \frac{17}{16}}$$

Take the square root of 289.

$$x=\frac{-\left(-\frac{17}{4}\right)±17}{2\times \frac{17}{16}}$$

The opposite of -\frac{17}{4} is \frac{17}{4}.

$$x=\frac{\frac{17}{4}±17}{2\times \frac{17}{16}}$$

Multiply 2 times \frac{17}{16}.

$$x=\frac{\frac{17}{4}±17}{\frac{17}{8}}$$

Now solve the equation x=\frac{\frac{17}{4}±17}{\frac{17}{8}} when ± is plus. Add \frac{17}{4} to 17.

$$x=\frac{\frac{85}{4}}{\frac{17}{8}}$$

Divide \frac{85}{4} by \frac{17}{8} by multiplying \frac{85}{4} by the reciprocal of \frac{17}{8}.

$$x=10$$

Now solve the equation x=\frac{\frac{17}{4}±17}{\frac{17}{8}} when ± is minus. Subtract 17 from \frac{17}{4}.

$$x=-\frac{\frac{51}{4}}{\frac{17}{8}}$$

Divide -\frac{51}{4} by \frac{17}{8} by multiplying -\frac{51}{4} by the reciprocal of \frac{17}{8}.

$$x=-6$$

The equation is now solved.

$$x=10 x=-6$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9+\left(\frac{x}{4}+\frac{7}{2}\right)^{2}=85$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{7}{2} times \frac{2}{2}.

$$x^{2}-6x+9+\left(\frac{x}{4}+\frac{7\times 2}{4}\right)^{2}=85$$

Since \frac{x}{4} and \frac{7\times 2}{4} have the same denominator, add them by adding their numerators.

$$x^{2}-6x+9+\left(\frac{x+7\times 2}{4}\right)^{2}=85$$

Do the multiplications in x+7\times 2.

$$x^{2}-6x+9+\left(\frac{x+14}{4}\right)^{2}=85$$

To raise \frac{x+14}{4} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-6x+9+\frac{\left(x+14\right)^{2}}{4^{2}}=85$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-6x+9 times \frac{4^{2}}{4^{2}}.

$$\frac{\left(x^{2}-6x+9\right)\times 4^{2}}{4^{2}}+\frac{\left(x+14\right)^{2}}{4^{2}}=85$$

Since \frac{\left(x^{2}-6x+9\right)\times 4^{2}}{4^{2}} and \frac{\left(x+14\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-6x+9\right)\times 4^{2}+\left(x+14\right)^{2}}{4^{2}}=85$$

Do the multiplications in \left(x^{2}-6x+9\right)\times 4^{2}+\left(x+14\right)^{2}.

$$\frac{16x^{2}-96x+144+x^{2}+28x+196}{4^{2}}=85$$

Combine like terms in 16x^{2}-96x+144+x^{2}+28x+196.

$$\frac{17x^{2}-68x+340}{4^{2}}=85$$

Calculate 4 to the power of 2 and get 16.

$$\frac{17x^{2}-68x+340}{16}=85$$

Divide each term of 17x^{2}-68x+340 by 16 to get \frac{17}{16}x^{2}-\frac{17}{4}x+\frac{85}{4}.

$$\frac{17}{16}x^{2}-\frac{17}{4}x+\frac{85}{4}=85$$

Subtract \frac{85}{4} from both sides.

$$\frac{17}{16}x^{2}-\frac{17}{4}x=85-\frac{85}{4}$$

Subtract \frac{85}{4} from 85 to get \frac{255}{4}.

$$\frac{17}{16}x^{2}-\frac{17}{4}x=\frac{255}{4}$$

Divide both sides of the equation by \frac{17}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{17}{16}x^{2}-\frac{17}{4}x}{\frac{17}{16}}=\frac{\frac{255}{4}}{\frac{17}{16}}$$

Dividing by \frac{17}{16} undoes the multiplication by \frac{17}{16}.

$$x^{2}+\left(-\frac{\frac{17}{4}}{\frac{17}{16}}\right)x=\frac{\frac{255}{4}}{\frac{17}{16}}$$

Divide -\frac{17}{4} by \frac{17}{16} by multiplying -\frac{17}{4} by the reciprocal of \frac{17}{16}.

$$x^{2}-4x=\frac{\frac{255}{4}}{\frac{17}{16}}$$

Divide \frac{255}{4} by \frac{17}{16} by multiplying \frac{255}{4} by the reciprocal of \frac{17}{16}.

$$x^{2}-4x=60$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=60+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=60+4$$

Add 60 to 4.

$$x^{2}-4x+4=64$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{64}$$

Simplify.

$$x-2=8 x-2=-8$$

Add 2 to both sides of the equation.

$$x=10 x=-6$$