Solve 3-(1/4x^2+x+2)=x-2

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

Solve for x

$$x=2\sqrt{7}-4\approx 1.291502622$$

$$x=-2\sqrt{7}-4\approx -9.291502622$$

Steps Using the Quadratic Formula

To find the opposite of \frac{1}{4}x^{2}+x+2, find the opposite of each term.

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

Subtract 2 from 3 to get 1.

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

Subtract x from both sides.

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

Combine -x and -x to get -2x.

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

Add 2 to both sides.

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

Add 1 and 2 to get 3.

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

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.

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

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

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

Square -2.

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

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

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

Add 4 to 3.

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

The opposite of -2 is 2.

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

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

$$x=\frac{2±\sqrt{7}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{2±\sqrt{7}}{-\frac{1}{2}} when ± is plus. Add 2 to \sqrt{7}.

$$x=\frac{\sqrt{7}+2}{-\frac{1}{2}}$$

Divide 2+\sqrt{7} by -\frac{1}{2} by multiplying 2+\sqrt{7} by the reciprocal of -\frac{1}{2}.

$$x=-2\sqrt{7}-4$$

Now solve the equation x=\frac{2±\sqrt{7}}{-\frac{1}{2}} when ± is minus. Subtract \sqrt{7} from 2.

$$x=\frac{2-\sqrt{7}}{-\frac{1}{2}}$$

Divide 2-\sqrt{7} by -\frac{1}{2} by multiplying 2-\sqrt{7} by the reciprocal of -\frac{1}{2}.

$$x=2\sqrt{7}-4$$

The equation is now solved.

$$x=-2\sqrt{7}-4 x=2\sqrt{7}-4$$

Steps for Completing the Square

To find the opposite of \frac{1}{4}x^{2}+x+2, find the opposite of each term.

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

Subtract 2 from 3 to get 1.

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

Subtract x from both sides.

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

Combine -x and -x to get -2x.

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

Subtract 1 from both sides.

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

Subtract 1 from -2 to get -3.

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

Multiply both sides by -4.

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

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

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

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

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

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

$$x^{2}+8x=12$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=12+4^{2}$$

Square 4.

$$x^{2}+8x+16=12+16$$

Add 12 to 16.

$$x^{2}+8x+16=28$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=28$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{28}$$

Simplify.

$$x+4=2\sqrt{7} x+4=-2\sqrt{7}$$

Subtract 4 from both sides of the equation.

$$x=2\sqrt{7}-4 x=-2\sqrt{7}-4$$