Solve 4-x/2=-1/2x^2+5/2x-2/4

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

Solve for x

$$x=4$$

$$x=5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4, the least common multiple of 2,4.

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

Use the distributive property to multiply 2 by 4-x.

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

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

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

Subtract \frac{5}{2}x from both sides.

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

Combine -2x and -\frac{5}{2}x to get -\frac{9}{2}x.

$$8-\frac{9}{2}x+\frac{1}{2}x^{2}=-2$$

Add 2 to both sides.

$$8-\frac{9}{2}x+\frac{1}{2}x^{2}+2=0$$

Add 8 and 2 to get 10.

$$10-\frac{9}{2}x+\frac{1}{2}x^{2}=0$$

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.

$$\frac{1}{2}x^{2}-\frac{9}{2}x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{9}{2} for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\left(-\frac{9}{2}\right)^{2}-4\times \frac{1}{2}\times 10}}{2\times \frac{1}{2}}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-4\times \frac{1}{2}\times 10}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-2\times 10}}{2\times \frac{1}{2}}$$

Multiply -2 times 10.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-20}}{2\times \frac{1}{2}}$$

Add \frac{81}{4} to -20.

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

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

$$x=\frac{-\left(-\frac{9}{2}\right)±\frac{1}{2}}{2\times \frac{1}{2}}$$

The opposite of -\frac{9}{2} is \frac{9}{2}.

$$x=\frac{\frac{9}{2}±\frac{1}{2}}{2\times \frac{1}{2}}$$

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

$$x=\frac{\frac{9}{2}±\frac{1}{2}}{1}$$

Now solve the equation x=\frac{\frac{9}{2}±\frac{1}{2}}{1} when ± is plus. Add \frac{9}{2} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 5 by 1.

$$x=5$$

Now solve the equation x=\frac{\frac{9}{2}±\frac{1}{2}}{1} when ± is minus. Subtract \frac{1}{2} from \frac{9}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 4 by 1.

$$x=4$$

The equation is now solved.

$$x=5 x=4$$

Steps for Completing the Square

Multiply both sides of the equation by 4, the least common multiple of 2,4.

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

Use the distributive property to multiply 2 by 4-x.

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

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

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

Subtract \frac{5}{2}x from both sides.

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

Combine -2x and -\frac{5}{2}x to get -\frac{9}{2}x.

$$8-\frac{9}{2}x+\frac{1}{2}x^{2}=-2$$

Subtract 8 from both sides.

$$-\frac{9}{2}x+\frac{1}{2}x^{2}=-2-8$$

Subtract 8 from -2 to get -10.

$$-\frac{9}{2}x+\frac{1}{2}x^{2}=-10$$

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.

$$\frac{1}{2}x^{2}-\frac{9}{2}x=-10$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-\frac{9}{2}x}{\frac{1}{2}}=-\frac{10}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\left(-\frac{\frac{9}{2}}{\frac{1}{2}}\right)x=-\frac{10}{\frac{1}{2}}$$

Divide -\frac{9}{2} by \frac{1}{2} by multiplying -\frac{9}{2} by the reciprocal of \frac{1}{2}.

$$x^{2}-9x=-\frac{10}{\frac{1}{2}}$$

Divide -10 by \frac{1}{2} by multiplying -10 by the reciprocal of \frac{1}{2}.

$$x^{2}-9x=-20$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-20+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{1}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{1}{2} x-\frac{9}{2}=-\frac{1}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=5 x=4$$