Solve n7-4(y+3)-(y+3)^2=0

$$n 7 - 4 ( y + 3 ) - ( y + 3 ) ^ { 2 } = 0$$

Solve for n_7

$$n_{7}=\left(y+3\right)\left(y+7\right)$$

Solution Steps

Use the distributive property to multiply -4 by y+3.

$$n_{7}-4y-12-\left(y+3\right)^{2}=0$$

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

$$n_{7}-4y-12-\left(y^{2}+6y+9\right)=0$$

To find the opposite of y^{2}+6y+9, find the opposite of each term.

$$n_{7}-4y-12-y^{2}-6y-9=0$$

Combine -4y and -6y to get -10y.

$$n_{7}-10y-12-y^{2}-9=0$$

Subtract 9 from -12 to get -21.

$$n_{7}-10y-21-y^{2}=0$$

Add 10y to both sides. Anything plus zero gives itself.

$$n_{7}-21-y^{2}=10y$$

Add 21 to both sides.

$$n_{7}-y^{2}=10y+21$$

Add y^{2} to both sides.

$$n_{7}=10y+21+y^{2}$$

Solve for y (complex solution)

$$y=-\left(\sqrt{n_{7}+4}+5\right)$$

$$y=\sqrt{n_{7}+4}-5$$

Solve for y

$$y=\sqrt{n_{7}+4}-5$$

$$y=-\left(\sqrt{n_{7}+4}+5\right)\text{, }n_{7}\geq -4$$