Solve left(5-2yright)^2+y^2=10

$${ \left(5-2y \right) }^{ 2 } + { y }^{ 2 } =10$$

Solve for y

$$y=1$$

$$y=3$$

Steps Using Factoring By Grouping

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

$$25-20y+4y^{2}+y^{2}=10$$

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

$$25-20y+5y^{2}=10$$

Subtract 10 from both sides.

$$25-20y+5y^{2}-10=0$$

Subtract 10 from 25 to get 15.

$$15-20y+5y^{2}=0$$

Divide both sides by 5.

$$3-4y+y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-4y+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite y^{2}-4y+3 as \left(y^{2}-3y\right)+\left(-y+3\right).

$$\left(y^{2}-3y\right)+\left(-y+3\right)$$

Factor out y in the first and -1 in the second group.

$$y\left(y-3\right)-\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(y-1\right)$$

To find equation solutions, solve y-3=0 and y-1=0.

$$y=3 y=1$$

Steps Using the Quadratic Formula

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

$$25-20y+4y^{2}+y^{2}=10$$

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

$$25-20y+5y^{2}=10$$

Subtract 10 from both sides.

$$25-20y+5y^{2}-10=0$$

Subtract 10 from 25 to get 15.

$$15-20y+5y^{2}=0$$

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.

$$5y^{2}-20y+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -20 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 5\times 15}}{2\times 5}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 5\times 15}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-20\right)±\sqrt{400-20\times 15}}{2\times 5}$$

Multiply -20 times 15.

$$y=\frac{-\left(-20\right)±\sqrt{400-300}}{2\times 5}$$

Add 400 to -300.

$$y=\frac{-\left(-20\right)±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$y=\frac{-\left(-20\right)±10}{2\times 5}$$

The opposite of -20 is 20.

$$y=\frac{20±10}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{20±10}{10}$$

Now solve the equation y=\frac{20±10}{10} when ± is plus. Add 20 to 10.

$$y=\frac{30}{10}$$

Divide 30 by 10.

$$y=3$$

Now solve the equation y=\frac{20±10}{10} when ± is minus. Subtract 10 from 20.

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

Divide 10 by 10.

$$y=1$$

The equation is now solved.

$$y=3 y=1$$

Steps for Completing the Square

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

$$25-20y+4y^{2}+y^{2}=10$$

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

$$25-20y+5y^{2}=10$$

Subtract 25 from both sides.

$$-20y+5y^{2}=10-25$$

Subtract 25 from 10 to get -15.

$$-20y+5y^{2}=-15$$

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.

$$5y^{2}-20y=-15$$

Divide both sides by 5.

$$\frac{5y^{2}-20y}{5}=-\frac{15}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\left(-\frac{20}{5}\right)y=-\frac{15}{5}$$

Divide -20 by 5.

$$y^{2}-4y=-\frac{15}{5}$$

Divide -15 by 5.

$$y^{2}-4y=-3$$

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.

$$y^{2}-4y+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$y^{2}-4y+4=-3+4$$

Add -3 to 4.

$$y^{2}-4y+4=1$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-2=1 y-2=-1$$

Add 2 to both sides of the equation.

$$y=3 y=1$$