Solve x^2+(-1/2x+2,5)^2=5

$$x ^ { 2 } + ( - \frac { 1 } { 2 } x + 2,5 ) ^ { 2 } = 5$$

Solve for x

$$x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{1}{2}x+2,5\right)^{2}.

$$x^{2}+\frac{1}{4}x^{2}-2,5x+6,25=5$$

Combine x^{2} and \frac{1}{4}x^{2} to get \frac{5}{4}x^{2}.

$$\frac{5}{4}x^{2}-2,5x+6,25=5$$

Subtract 5 from both sides.

$$\frac{5}{4}x^{2}-2,5x+6,25-5=0$$

Subtract 5 from 6,25 to get 1,25.

$$\frac{5}{4}x^{2}-2,5x+1,25=0$$

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.

$$\frac{5}{4}x^{2}-2,5x+\frac{5}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{4} for a, -2,5 for b, and \frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2,5\right)±\sqrt{\left(-2,5\right)^{2}-4\times \frac{5}{4}\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

Square -2,5 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-2,5\right)±\sqrt{6,25-4\times \frac{5}{4}\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

Multiply -4 times \frac{5}{4}.

$$x=\frac{-\left(-2,5\right)±\sqrt{6,25-5\times \frac{5}{4}}}{2\times \frac{5}{4}}$$

Multiply -5 times \frac{5}{4}.

$$x=\frac{-\left(-2,5\right)±\sqrt{\frac{25-25}{4}}}{2\times \frac{5}{4}}$$

Add 6,25 to -\frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-2,5\right)±\sqrt{0}}{2\times \frac{5}{4}}$$

Take the square root of 0.

$$x=-\frac{-2,5}{2\times \frac{5}{4}}$$

The opposite of -2,5 is 2,5.

$$x=\frac{2,5}{2\times \frac{5}{4}}$$

Multiply 2 times \frac{5}{4}.

$$x=\frac{2,5}{\frac{5}{2}}$$

Divide 2,5 by \frac{5}{2} by multiplying 2,5 by the reciprocal of \frac{5}{2}.

$$x=1$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{1}{2}x+2,5\right)^{2}.

$$x^{2}+\frac{1}{4}x^{2}-2,5x+6,25=5$$

Combine x^{2} and \frac{1}{4}x^{2} to get \frac{5}{4}x^{2}.

$$\frac{5}{4}x^{2}-2,5x+6,25=5$$

Subtract 6,25 from both sides.

$$\frac{5}{4}x^{2}-2,5x=5-6,25$$

Subtract 6,25 from 5 to get -1,25.

$$\frac{5}{4}x^{2}-2,5x=-1,25$$

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{4}x^{2}-2,5x}{\frac{5}{4}}=-\frac{1,25}{\frac{5}{4}}$$

Dividing by \frac{5}{4} undoes the multiplication by \frac{5}{4}.

$$x^{2}+\left(-\frac{2,5}{\frac{5}{4}}\right)x=-\frac{1,25}{\frac{5}{4}}$$

Divide -2,5 by \frac{5}{4} by multiplying -2,5 by the reciprocal of \frac{5}{4}.

$$x^{2}-2x=-\frac{1,25}{\frac{5}{4}}$$

Divide -1,25 by \frac{5}{4} by multiplying -1,25 by the reciprocal of \frac{5}{4}.

$$x^{2}-2x=-1$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-1+1$$

Add -1 to 1.

$$x^{2}-2x+1=0$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$