Solve y^2-2y-18=2(y-3)

$$y ^ { 2 } - 2 y - 18 = 2 ( y - 3 )$$

Solve for y

$$y=-2$$

$$y=6$$

Steps Using Factoring

Use the distributive property to multiply 2 by y-3.

$$y^{2}-2y-18=2y-6$$

Subtract 2y from both sides.

$$y^{2}-2y-18-2y=-6$$

Combine -2y and -2y to get -4y.

$$y^{2}-4y-18=-6$$

Add 6 to both sides.

$$y^{2}-4y-18+6=0$$

Add -18 and 6 to get -12.

$$y^{2}-4y-12=0$$

To solve the equation, factor y^{2}-4y-12 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.

$$a+b=-4 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

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

$$\left(y-6\right)\left(y+2\right)$$

To find equation solutions, solve y-6=0 and y+2=0.

$$y=6 y=-2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2 by y-3.

$$y^{2}-2y-18=2y-6$$

Subtract 2y from both sides.

$$y^{2}-2y-18-2y=-6$$

Combine -2y and -2y to get -4y.

$$y^{2}-4y-18=-6$$

Add 6 to both sides.

$$y^{2}-4y-18+6=0$$

Add -18 and 6 to get -12.

$$y^{2}-4y-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-12. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite y^{2}-4y-12 as \left(y^{2}-6y\right)+\left(2y-12\right).

$$\left(y^{2}-6y\right)+\left(2y-12\right)$$

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

$$y\left(y-6\right)+2\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(y+2\right)$$

To find equation solutions, solve y-6=0 and y+2=0.

$$y=6 y=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by y-3.

$$y^{2}-2y-18=2y-6$$

Subtract 2y from both sides.

$$y^{2}-2y-18-2y=-6$$

Combine -2y and -2y to get -4y.

$$y^{2}-4y-18=-6$$

Add 6 to both sides.

$$y^{2}-4y-18+6=0$$

Add -18 and 6 to get -12.

$$y^{2}-4y-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$y=\frac{-\left(-4\right)±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$y=\frac{-\left(-4\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$y=\frac{-\left(-4\right)±8}{2}$$

The opposite of -4 is 4.

$$y=\frac{4±8}{2}$$

Now solve the equation y=\frac{4±8}{2} when ± is plus. Add 4 to 8.

$$y=\frac{12}{2}$$

Divide 12 by 2.

$$y=6$$

Now solve the equation y=\frac{4±8}{2} when ± is minus. Subtract 8 from 4.

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

Divide -4 by 2.

$$y=-2$$

The equation is now solved.

$$y=6 y=-2$$

Steps for Completing the Square

Use the distributive property to multiply 2 by y-3.

$$y^{2}-2y-18=2y-6$$

Subtract 2y from both sides.

$$y^{2}-2y-18-2y=-6$$

Combine -2y and -2y to get -4y.

$$y^{2}-4y-18=-6$$

Add 18 to both sides.

$$y^{2}-4y=-6+18$$

Add -6 and 18 to get 12.

$$y^{2}-4y=12$$

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.

$$y^{2}-4y+\left(-2\right)^{2}=12+\left(-2\right)^{2}$$

Square -2.

$$y^{2}-4y+4=12+4$$

Add 12 to 4.

$$y^{2}-4y+4=16$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(y-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$y-2=4 y-2=-4$$

Add 2 to both sides of the equation.

$$y=6 y=-2$$