Solve -3/4x+8=-1/8x^2+x+3

$$- \frac{ 3 }{ 4 } x+8=- \frac{ 1 }{ 8 } { x }^{ 2 } +x+3$$

Solve for x

$$x=4$$

$$x=10$$

Steps Using the Quadratic Formula

Add \frac{1}{8}x^{2} to both sides.

$$-\frac{3}{4}x+8+\frac{1}{8}x^{2}=x+3$$

Subtract x from both sides.

$$-\frac{3}{4}x+8+\frac{1}{8}x^{2}-x=3$$

Combine -\frac{3}{4}x and -x to get -\frac{7}{4}x.

$$-\frac{7}{4}x+8+\frac{1}{8}x^{2}=3$$

Subtract 3 from both sides.

$$-\frac{7}{4}x+8+\frac{1}{8}x^{2}-3=0$$

Subtract 3 from 8 to get 5.

$$-\frac{7}{4}x+5+\frac{1}{8}x^{2}=0$$

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.

$$\frac{1}{8}x^{2}-\frac{7}{4}x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{8} for a, -\frac{7}{4} for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{7}{4}\right)±\sqrt{\left(-\frac{7}{4}\right)^{2}-4\times \frac{1}{8}\times 5}}{2\times \frac{1}{8}}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{7}{4}\right)±\sqrt{\frac{49}{16}-4\times \frac{1}{8}\times 5}}{2\times \frac{1}{8}}$$

Multiply -4 times \frac{1}{8}.

$$x=\frac{-\left(-\frac{7}{4}\right)±\sqrt{\frac{49}{16}-\frac{1}{2}\times 5}}{2\times \frac{1}{8}}$$

Multiply -\frac{1}{2} times 5.

$$x=\frac{-\left(-\frac{7}{4}\right)±\sqrt{\frac{49}{16}-\frac{5}{2}}}{2\times \frac{1}{8}}$$

Add \frac{49}{16} to -\frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{7}{4}\right)±\sqrt{\frac{9}{16}}}{2\times \frac{1}{8}}$$

Take the square root of \frac{9}{16}.

$$x=\frac{-\left(-\frac{7}{4}\right)±\frac{3}{4}}{2\times \frac{1}{8}}$$

The opposite of -\frac{7}{4} is \frac{7}{4}.

$$x=\frac{\frac{7}{4}±\frac{3}{4}}{2\times \frac{1}{8}}$$

Multiply 2 times \frac{1}{8}.

$$x=\frac{\frac{7}{4}±\frac{3}{4}}{\frac{1}{4}}$$

Now solve the equation x=\frac{\frac{7}{4}±\frac{3}{4}}{\frac{1}{4}} when ± is plus. Add \frac{7}{4} to \frac{3}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{5}{2}}{\frac{1}{4}}$$

Divide \frac{5}{2} by \frac{1}{4} by multiplying \frac{5}{2} by the reciprocal of \frac{1}{4}.

$$x=10$$

Now solve the equation x=\frac{\frac{7}{4}±\frac{3}{4}}{\frac{1}{4}} when ± is minus. Subtract \frac{3}{4} from \frac{7}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1}{\frac{1}{4}}$$

Divide 1 by \frac{1}{4} by multiplying 1 by the reciprocal of \frac{1}{4}.

$$x=4$$

The equation is now solved.

$$x=10 x=4$$

Steps for Completing the Square

Add \frac{1}{8}x^{2} to both sides.

$$-\frac{3}{4}x+8+\frac{1}{8}x^{2}=x+3$$

Subtract x from both sides.

$$-\frac{3}{4}x+8+\frac{1}{8}x^{2}-x=3$$

Combine -\frac{3}{4}x and -x to get -\frac{7}{4}x.

$$-\frac{7}{4}x+8+\frac{1}{8}x^{2}=3$$

Subtract 8 from both sides.

$$-\frac{7}{4}x+\frac{1}{8}x^{2}=3-8$$

Subtract 8 from 3 to get -5.

$$-\frac{7}{4}x+\frac{1}{8}x^{2}=-5$$

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.

$$\frac{1}{8}x^{2}-\frac{7}{4}x=-5$$

Multiply both sides by 8.

$$\frac{\frac{1}{8}x^{2}-\frac{7}{4}x}{\frac{1}{8}}=-\frac{5}{\frac{1}{8}}$$

Dividing by \frac{1}{8} undoes the multiplication by \frac{1}{8}.

$$x^{2}+\left(-\frac{\frac{7}{4}}{\frac{1}{8}}\right)x=-\frac{5}{\frac{1}{8}}$$

Divide -\frac{7}{4} by \frac{1}{8} by multiplying -\frac{7}{4} by the reciprocal of \frac{1}{8}.

$$x^{2}-14x=-\frac{5}{\frac{1}{8}}$$

Divide -5 by \frac{1}{8} by multiplying -5 by the reciprocal of \frac{1}{8}.

$$x^{2}-14x=-40$$

Divide -14, the coefficient of the x term, by 2 to get -7. Then add the square of -7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-14x+\left(-7\right)^{2}=-40+\left(-7\right)^{2}$$

Square -7.

$$x^{2}-14x+49=-40+49$$

Add -40 to 49.

$$x^{2}-14x+49=9$$

Factor x^{2}-14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-7\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-7\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-7=3 x-7=-3$$

Add 7 to both sides of the equation.

$$x=10 x=4$$