Solve (2-x)^2+(1.5+0.5)^2=2.5^2

$$( 2 - x ) ^ { 2 } + ( 1.5 + 0.5 ) ^ { 2 } = 2.5 ^ { 2 }$$

Solve for x

$$x=0.5$$

$$x=3.5$$

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}.

$$4-4x+x^{2}+\left(1.5+0.5\right)^{2}=2.5^{2}$$

Add 1.5 and 0.5 to get 2.

$$4-4x+x^{2}+2^{2}=2.5^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4-4x+x^{2}+4=2.5^{2}$$

Add 4 and 4 to get 8.

$$8-4x+x^{2}=2.5^{2}$$

Calculate 2.5 to the power of 2 and get 6.25.

$$8-4x+x^{2}=6.25$$

Subtract 6.25 from both sides.

$$8-4x+x^{2}-6.25=0$$

Subtract 6.25 from 8 to get 1.75.

$$1.75-4x+x^{2}=0$$

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.

$$x^{2}-4x+1.75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 1.75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1.75}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 1.75}}{2}$$

Multiply -4 times 1.75.

$$x=\frac{-\left(-4\right)±\sqrt{16-7}}{2}$$

Add 16 to -7.

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

Take the square root of 9.

$$x=\frac{-\left(-4\right)±3}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±3}{2}$$

Now solve the equation x=\frac{4±3}{2} when ± is plus. Add 4 to 3.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{4±3}{2} when ± is minus. Subtract 3 from 4.

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

The equation is now solved.

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

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}.

$$4-4x+x^{2}+\left(1.5+0.5\right)^{2}=2.5^{2}$$

Add 1.5 and 0.5 to get 2.

$$4-4x+x^{2}+2^{2}=2.5^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4-4x+x^{2}+4=2.5^{2}$$

Add 4 and 4 to get 8.

$$8-4x+x^{2}=2.5^{2}$$

Calculate 2.5 to the power of 2 and get 6.25.

$$8-4x+x^{2}=6.25$$

Subtract 8 from both sides.

$$-4x+x^{2}=6.25-8$$

Subtract 8 from 6.25 to get -1.75.

$$-4x+x^{2}=-1.75$$

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.

$$x^{2}-4x=-1.75$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-1.75+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-1.75+4$$

Add -1.75 to 4.

$$x^{2}-4x+4=2.25$$

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}.

$$\left(x-2\right)^{2}=2.25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{2.25}$$

Simplify.

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

Add 2 to both sides of the equation.

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