Solve y^2+6y+15=3-7y

$$y ^ { 2 } + 6 y + 15 = 3 - 7 y$$

Solve for y

$$y=-12$$

$$y=-1$$

Steps Using Factoring

Subtract 3 from both sides.

$$y^{2}+6y+15-3=-7y$$

Subtract 3 from 15 to get 12.

$$y^{2}+6y+12=-7y$$

Add 7y to both sides.

$$y^{2}+6y+12+7y=0$$

Combine 6y and 7y to get 13y.

$$y^{2}+13y+12=0$$

To solve the equation, factor y^{2}+13y+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.

$$a+b=13 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=1 b=12$$

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

$$\left(y+1\right)\left(y+12\right)$$

To find equation solutions, solve y+1=0 and y+12=0.

$$y=-1 y=-12$$

Steps Using Factoring By Grouping

Subtract 3 from both sides.

$$y^{2}+6y+15-3=-7y$$

Subtract 3 from 15 to get 12.

$$y^{2}+6y+12=-7y$$

Add 7y to both sides.

$$y^{2}+6y+12+7y=0$$

Combine 6y and 7y to get 13y.

$$y^{2}+13y+12=0$$

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.

$$a+b=13 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=1 b=12$$

Rewrite y^{2}+13y+12 as \left(y^{2}+y\right)+\left(12y+12\right).

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

Factor out y in the first and 12 in the second group.

$$y\left(y+1\right)+12\left(y+1\right)$$

Factor out common term y+1 by using distributive property.

$$\left(y+1\right)\left(y+12\right)$$

To find equation solutions, solve y+1=0 and y+12=0.

$$y=-1 y=-12$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

$$y^{2}+6y+15-3=-7y$$

Subtract 3 from 15 to get 12.

$$y^{2}+6y+12=-7y$$

Add 7y to both sides.

$$y^{2}+6y+12+7y=0$$

Combine 6y and 7y to get 13y.

$$y^{2}+13y+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 13 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-13±\sqrt{13^{2}-4\times 12}}{2}$$

Square 13.

$$y=\frac{-13±\sqrt{169-4\times 12}}{2}$$

Multiply -4 times 12.

$$y=\frac{-13±\sqrt{169-48}}{2}$$

Add 169 to -48.

$$y=\frac{-13±\sqrt{121}}{2}$$

Take the square root of 121.

$$y=\frac{-13±11}{2}$$

Now solve the equation y=\frac{-13±11}{2} when ± is plus. Add -13 to 11.

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

Divide -2 by 2.

$$y=-1$$

Now solve the equation y=\frac{-13±11}{2} when ± is minus. Subtract 11 from -13.

$$y=-\frac{24}{2}$$

Divide -24 by 2.

$$y=-12$$

The equation is now solved.

$$y=-1 y=-12$$

Steps for Completing the Square

Add 7y to both sides.

$$y^{2}+6y+15+7y=3$$

Combine 6y and 7y to get 13y.

$$y^{2}+13y+15=3$$

Subtract 15 from both sides.

$$y^{2}+13y=3-15$$

Subtract 15 from 3 to get -12.

$$y^{2}+13y=-12$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+13y+\left(\frac{13}{2}\right)^{2}=-12+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+13y+\frac{169}{4}=-12+\frac{169}{4}$$

Add -12 to \frac{169}{4}.

$$y^{2}+13y+\frac{169}{4}=\frac{121}{4}$$

Factor y^{2}+13y+\frac{169}{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}.

$$\left(y+\frac{13}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{13}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$y+\frac{13}{2}=\frac{11}{2} y+\frac{13}{2}=-\frac{11}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$y=-1 y=-12$$