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

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

Solve for x

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

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

Steps Using the Quadratic Formula

Add \frac{1}{4}x^{2} to both sides.

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

Subtract \frac{3}{2}x from both sides.

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

Combine -\frac{1}{2}x and -\frac{3}{2}x to get -2x.

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

Subtract 4 from both sides.

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

Subtract 4 from 1 to get -3.

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

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.

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

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}.

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

Square -2.

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

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

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

Multiply -1 times -3.

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

Add 4 to 3.

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

The opposite of -2 is 2.

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

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}.

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

The equation is now solved.

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

Steps for Completing the Square

Add \frac{1}{4}x^{2} to both sides.

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

Subtract \frac{3}{2}x from both sides.

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

Combine -\frac{1}{2}x and -\frac{3}{2}x to get -2x.

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

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

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.

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

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.

$$x^{2}-8x+\left(-4\right)^{2}=12+\left(-4\right)^{2}$$

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.

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

Add 4 to both sides of the equation.

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