Solve (2y-5)^2+y^2=5

$$( 2 y - 5 ) ^ { 2 } + y ^ { 2 } = 5$$

Solve for y

$$y=2$$

Steps Using Factoring By Grouping

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

$$4y^{2}-20y+25+y^{2}=5$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$5y^{2}-20y+25=5$$

Subtract 5 from both sides.

$$5y^{2}-20y+25-5=0$$

Subtract 5 from 25 to get 20.

$$5y^{2}-20y+20=0$$

Divide both sides by 5.

$$y^{2}-4y+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite y^{2}-4y+4 as \left(y^{2}-2y\right)+\left(-2y+4\right).

$$\left(y^{2}-2y\right)+\left(-2y+4\right)$$

Factor out y in the first and -2 in the second group.

$$y\left(y-2\right)-2\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(y-2\right)$$

Rewrite as a binomial square.

$$\left(y-2\right)^{2}$$

To find equation solution, solve y-2=0.

$$y=2$$

Steps Using the Quadratic Formula

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

$$4y^{2}-20y+25+y^{2}=5$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$5y^{2}-20y+25=5$$

Subtract 5 from both sides.

$$5y^{2}-20y+25-5=0$$

Subtract 5 from 25 to get 20.

$$5y^{2}-20y+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -20 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 5\times 20}}{2\times 5}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 5\times 20}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-20\right)±\sqrt{400-20\times 20}}{2\times 5}$$

Multiply -20 times 20.

$$y=\frac{-\left(-20\right)±\sqrt{400-400}}{2\times 5}$$

Add 400 to -400.

$$y=\frac{-\left(-20\right)±\sqrt{0}}{2\times 5}$$

Take the square root of 0.

$$y=-\frac{-20}{2\times 5}$$

The opposite of -20 is 20.

$$y=\frac{20}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{20}{10}$$

Divide 20 by 10.

$$y=2$$

Steps for Completing the Square

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

$$4y^{2}-20y+25+y^{2}=5$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$5y^{2}-20y+25=5$$

Subtract 25 from both sides.

$$5y^{2}-20y=5-25$$

Subtract 25 from 5 to get -20.

$$5y^{2}-20y=-20$$

Divide both sides by 5.

$$\frac{5y^{2}-20y}{5}=-\frac{20}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\left(-\frac{20}{5}\right)y=-\frac{20}{5}$$

Divide -20 by 5.

$$y^{2}-4y=-\frac{20}{5}$$

Divide -20 by 5.

$$y^{2}-4y=-4$$

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.

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

Square -2.

$$y^{2}-4y+4=-4+4$$

Add -4 to 4.

$$y^{2}-4y+4=0$$

Factor y^{2}-4y+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-2=0 y-2=0$$

Add 2 to both sides of the equation.

$$y=2 y=2$$

The equation is now solved. Solutions are the same.

$$y=2$$