Solve x=x^2+3/4

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

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}.

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

Subtract \frac{1}{4}x^{2} from both sides.

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

Subtract \frac{3}{4} from both sides.

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

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}.

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

Square 1.

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

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

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

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

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

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

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

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}.

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

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

$$x=1$$

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

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

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

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

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

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

Subtract \frac{1}{4}x^{2} from both sides.

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

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.

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

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}.

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

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.

$$x=3 x=1$$