Solve (2-x)^2+x=10

$$( 2 - x ) ^ { 2 } + x = 10$$

Solve for x

$$x = \frac{\sqrt{33} + 3}{2} \approx 4.372281323$$

$$x=\frac{3-\sqrt{33}}{2}\approx -1.372281323$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.

$$4-4x+x^{2}+x=10$$

Combine -4x and x to get -3x.

$$4-3x+x^{2}=10$$

Subtract 10 from both sides.

$$4-3x+x^{2}-10=0$$

Subtract 10 from 4 to get -6.

$$-6-3x+x^{2}=0$$

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.

$$x^{2}-3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-6\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24}}{2}$$

Add 9 to 24.

$$x=\frac{-\left(-3\right)±\sqrt{33}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{33}}{2}$$

Now solve the equation x=\frac{3±\sqrt{33}}{2} when ± is plus. Add 3 to \sqrt{33}.

$$x=\frac{\sqrt{33}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{33}}{2} when ± is minus. Subtract \sqrt{33} from 3.

$$x=\frac{3-\sqrt{33}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+3}{2} x=\frac{3-\sqrt{33}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.

$$4-4x+x^{2}+x=10$$

Combine -4x and x to get -3x.

$$4-3x+x^{2}=10$$

Subtract 4 from both sides.

$$-3x+x^{2}=10-4$$

Subtract 4 from 10 to get 6.

$$-3x+x^{2}=6$$

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.

$$x^{2}-3x=6$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=6+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=6+\frac{9}{4}$$

Add 6 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{33}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{33}}{2} x-\frac{3}{2}=-\frac{\sqrt{33}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{33}+3}{2} x=\frac{3-\sqrt{33}}{2}$$