Solve 144+(y-2)^2=4(y+1)^2+36

$$144 + ( y - 2 ) ^ { 2 } = 4 ( y + 1 ) ^ { 2 } + 36$$

Solve for y

$$y=2\sqrt{10}-2\approx 4.32455532$$

$$y=-2\sqrt{10}-2\approx -8.32455532$$

Steps Using the Quadratic Formula

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

$$144+y^{2}-4y+4=4\left(y+1\right)^{2}+36$$

Add 144 and 4 to get 148.

$$148+y^{2}-4y=4\left(y+1\right)^{2}+36$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$148+y^{2}-4y=4\left(y^{2}+2y+1\right)+36$$

Use the distributive property to multiply 4 by y^{2}+2y+1.

$$148+y^{2}-4y=4y^{2}+8y+4+36$$

Add 4 and 36 to get 40.

$$148+y^{2}-4y=4y^{2}+8y+40$$

Subtract 4y^{2} from both sides.

$$148+y^{2}-4y-4y^{2}=8y+40$$

Combine y^{2} and -4y^{2} to get -3y^{2}.

$$148-3y^{2}-4y=8y+40$$

Subtract 8y from both sides.

$$148-3y^{2}-4y-8y=40$$

Combine -4y and -8y to get -12y.

$$148-3y^{2}-12y=40$$

Subtract 40 from both sides.

$$148-3y^{2}-12y-40=0$$

Subtract 40 from 148 to get 108.

$$108-3y^{2}-12y=0$$

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.

$$-3y^{2}-12y+108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -12 for b, and 108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-3\right)\times 108}}{2\left(-3\right)}$$

Square -12.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\left(-3\right)\times 108}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-\left(-12\right)±\sqrt{144+12\times 108}}{2\left(-3\right)}$$

Multiply 12 times 108.

$$y=\frac{-\left(-12\right)±\sqrt{144+1296}}{2\left(-3\right)}$$

Add 144 to 1296.

$$y=\frac{-\left(-12\right)±\sqrt{1440}}{2\left(-3\right)}$$

Take the square root of 1440.

$$y=\frac{-\left(-12\right)±12\sqrt{10}}{2\left(-3\right)}$$

The opposite of -12 is 12.

$$y=\frac{12±12\sqrt{10}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{12±12\sqrt{10}}{-6}$$

Now solve the equation y=\frac{12±12\sqrt{10}}{-6} when ± is plus. Add 12 to 12\sqrt{10}.

$$y=\frac{12\sqrt{10}+12}{-6}$$

Divide 12+12\sqrt{10} by -6.

$$y=-2\sqrt{10}-2$$

Now solve the equation y=\frac{12±12\sqrt{10}}{-6} when ± is minus. Subtract 12\sqrt{10} from 12.

$$y=\frac{12-12\sqrt{10}}{-6}$$

Divide 12-12\sqrt{10} by -6.

$$y=2\sqrt{10}-2$$

The equation is now solved.

$$y=-2\sqrt{10}-2 y=2\sqrt{10}-2$$

Steps for Completing the Square

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

$$144+y^{2}-4y+4=4\left(y+1\right)^{2}+36$$

Add 144 and 4 to get 148.

$$148+y^{2}-4y=4\left(y+1\right)^{2}+36$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$148+y^{2}-4y=4\left(y^{2}+2y+1\right)+36$$

Use the distributive property to multiply 4 by y^{2}+2y+1.

$$148+y^{2}-4y=4y^{2}+8y+4+36$$

Add 4 and 36 to get 40.

$$148+y^{2}-4y=4y^{2}+8y+40$$

Subtract 4y^{2} from both sides.

$$148+y^{2}-4y-4y^{2}=8y+40$$

Combine y^{2} and -4y^{2} to get -3y^{2}.

$$148-3y^{2}-4y=8y+40$$

Subtract 8y from both sides.

$$148-3y^{2}-4y-8y=40$$

Combine -4y and -8y to get -12y.

$$148-3y^{2}-12y=40$$

Subtract 148 from both sides.

$$-3y^{2}-12y=40-148$$

Subtract 148 from 40 to get -108.

$$-3y^{2}-12y=-108$$

Divide both sides by -3.

$$\frac{-3y^{2}-12y}{-3}=-\frac{108}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y^{2}+\left(-\frac{12}{-3}\right)y=-\frac{108}{-3}$$

Divide -12 by -3.

$$y^{2}+4y=-\frac{108}{-3}$$

Divide -108 by -3.

$$y^{2}+4y=36$$

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.

$$y^{2}+4y+2^{2}=36+2^{2}$$

Square 2.

$$y^{2}+4y+4=36+4$$

Add 36 to 4.

$$y^{2}+4y+4=40$$

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}.

$$\left(y+2\right)^{2}=40$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{40}$$

Simplify.

$$y+2=2\sqrt{10} y+2=-2\sqrt{10}$$

Subtract 2 from both sides of the equation.

$$y=2\sqrt{10}-2 y=-2\sqrt{10}-2$$