Solve 169=(5+7)^2+(y+3)^2

$$169 = ( 5 + 7 ) ^ { 2 } + ( y + 3 ) ^ { 2 }$$

Solve for y

$$y=2$$

$$y=-8$$

Steps Using Factoring

Add 5 and 7 to get 12.

$$169=12^{2}+\left(y+3\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$169=144+\left(y+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$169=144+y^{2}+6y+9$$

Add 144 and 9 to get 153.

$$169=153+y^{2}+6y$$

Swap sides so that all variable terms are on the left hand side.

$$153+y^{2}+6y=169$$

Subtract 169 from both sides.

$$153+y^{2}+6y-169=0$$

Subtract 169 from 153 to get -16.

$$-16+y^{2}+6y=0$$

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

$$y^{2}+6y-16=0$$

To solve the equation, factor y^{2}+6y-16 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=6 ab=-16$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-2\right)\left(y+8\right)$$

To find equation solutions, solve y-2=0 and y+8=0.

$$y=2 y=-8$$

Steps Using Factoring By Grouping

Add 5 and 7 to get 12.

$$169=12^{2}+\left(y+3\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$169=144+\left(y+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$169=144+y^{2}+6y+9$$

Add 144 and 9 to get 153.

$$169=153+y^{2}+6y$$

Swap sides so that all variable terms are on the left hand side.

$$153+y^{2}+6y=169$$

Subtract 169 from both sides.

$$153+y^{2}+6y-169=0$$

Subtract 169 from 153 to get -16.

$$-16+y^{2}+6y=0$$

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

$$y^{2}+6y-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-16. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-16\right)=-16$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

Rewrite y^{2}+6y-16 as \left(y^{2}-2y\right)+\left(8y-16\right).

$$\left(y^{2}-2y\right)+\left(8y-16\right)$$

Factor out y in the first and 8 in the second group.

$$y\left(y-2\right)+8\left(y-2\right)$$

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

$$\left(y-2\right)\left(y+8\right)$$

To find equation solutions, solve y-2=0 and y+8=0.

$$y=2 y=-8$$

Steps Using the Quadratic Formula

Add 5 and 7 to get 12.

$$169=12^{2}+\left(y+3\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$169=144+\left(y+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$169=144+y^{2}+6y+9$$

Add 144 and 9 to get 153.

$$169=153+y^{2}+6y$$

Swap sides so that all variable terms are on the left hand side.

$$153+y^{2}+6y=169$$

Subtract 169 from both sides.

$$153+y^{2}+6y-169=0$$

Subtract 169 from 153 to get -16.

$$-16+y^{2}+6y=0$$

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.

$$y^{2}+6y-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\left(-16\right)}}{2}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$y=\frac{-6±\sqrt{36+64}}{2}$$

Add 36 to 64.

$$y=\frac{-6±\sqrt{100}}{2}$$

Take the square root of 100.

$$y=\frac{-6±10}{2}$$

Now solve the equation y=\frac{-6±10}{2} when ± is plus. Add -6 to 10.

$$y=\frac{4}{2}$$

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{-6±10}{2} when ± is minus. Subtract 10 from -6.

$$y=-\frac{16}{2}$$

Divide -16 by 2.

$$y=-8$$

The equation is now solved.

$$y=2 y=-8$$

Steps for Completing the Square

Add 5 and 7 to get 12.

$$169=12^{2}+\left(y+3\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$169=144+\left(y+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$169=144+y^{2}+6y+9$$

Add 144 and 9 to get 153.

$$169=153+y^{2}+6y$$

Swap sides so that all variable terms are on the left hand side.

$$153+y^{2}+6y=169$$

Subtract 153 from both sides.

$$y^{2}+6y=169-153$$

Subtract 153 from 169 to get 16.

$$y^{2}+6y=16$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+6y+3^{2}=16+3^{2}$$

Square 3.

$$y^{2}+6y+9=16+9$$

Add 16 to 9.

$$y^{2}+6y+9=25$$

Factor y^{2}+6y+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+3\right)^{2}}=\sqrt{25}$$

Simplify.

$$y+3=5 y+3=-5$$

Subtract 3 from both sides of the equation.

$$y=2 y=-8$$