Solve x^2+3/4=x

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

Solve for x

$$x=1$$

$$x=3$$

Steps Using the Quadratic Formula

Divide each term of x^{2}+3 by 4 to get \frac{1}{4}x^{2}+\frac{3}{4}.

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

Subtract x from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -1 for b, and \frac{3}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

Add 1 to -\frac{3}{4}.

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

Take the square root of \frac{1}{4}.

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

The opposite of -1 is 1.

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

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

$$x=\frac{1±\frac{1}{2}}{\frac{1}{2}}$$

Now solve the equation x=\frac{1±\frac{1}{2}}{\frac{1}{2}} when ± is plus. Add 1 to \frac{1}{2}.

$$x=\frac{\frac{3}{2}}{\frac{1}{2}}$$

Divide \frac{3}{2} by \frac{1}{2} by multiplying \frac{3}{2} by the reciprocal of \frac{1}{2}.

$$x=3$$

Now solve the equation x=\frac{1±\frac{1}{2}}{\frac{1}{2}} when ± is minus. Subtract \frac{1}{2} from 1.

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

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

$$x=1$$

The equation is now solved.

$$x=3 x=1$$

Steps for Completing the Square

Divide each term of x^{2}+3 by 4 to get \frac{1}{4}x^{2}+\frac{3}{4}.

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

Subtract x from both sides.

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

Subtract \frac{3}{4} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by 4.

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

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$x^{2}+\left(-\frac{1}{\frac{1}{4}}\right)x=-\frac{\frac{3}{4}}{\frac{1}{4}}$$

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

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

Divide -\frac{3}{4} by \frac{1}{4} by multiplying -\frac{3}{4} by the reciprocal of \frac{1}{4}.

$$x^{2}-4x=-3$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

$$x^{2}-4x+4=1$$

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}.

$$\left(x-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$