Solve (x+3)^2=2x(x+7)

$$( x + 3 ) ^ { 2 } = 2 x ( x + 7 )$$

Solve for x

$$x=-9$$

$$x=1$$

Steps Using Factoring By Grouping

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

$$x^{2}+6x+9=2x\left(x+7\right)$$

Use the distributive property to multiply 2x by x+7.

$$x^{2}+6x+9=2x^{2}+14x$$

Subtract 2x^{2} from both sides.

$$x^{2}+6x+9-2x^{2}=14x$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+6x+9=14x$$

Subtract 14x from both sides.

$$-x^{2}+6x+9-14x=0$$

Combine 6x and -14x to get -8x.

$$-x^{2}-8x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-9=-9$$

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 -9.

$$1,-9 3,-3$$

Calculate the sum for each pair.

$$1-9=-8 3-3=0$$

The solution is the pair that gives sum -8.

$$a=1 b=-9$$

Rewrite -x^{2}-8x+9 as \left(-x^{2}+x\right)+\left(-9x+9\right).

$$\left(-x^{2}+x\right)+\left(-9x+9\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(-x+1\right)+9\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+9\right)$$

To find equation solutions, solve -x+1=0 and x+9=0.

$$x=1 x=-9$$

Steps Using the Quadratic Formula

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

$$x^{2}+6x+9=2x\left(x+7\right)$$

Use the distributive property to multiply 2x by x+7.

$$x^{2}+6x+9=2x^{2}+14x$$

Subtract 2x^{2} from both sides.

$$x^{2}+6x+9-2x^{2}=14x$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+6x+9=14x$$

Subtract 14x from both sides.

$$-x^{2}+6x+9-14x=0$$

Combine 6x and -14x to get -8x.

$$-x^{2}-8x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-8\right)±\sqrt{64+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-\left(-8\right)±\sqrt{64+36}}{2\left(-1\right)}$$

Add 64 to 36.

$$x=\frac{-\left(-8\right)±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$x=\frac{-\left(-8\right)±10}{2\left(-1\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{8±10}{-2}$$

Now solve the equation x=\frac{8±10}{-2} when ± is plus. Add 8 to 10.

$$x=\frac{18}{-2}$$

Divide 18 by -2.

$$x=-9$$

Now solve the equation x=\frac{8±10}{-2} when ± is minus. Subtract 10 from 8.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-9 x=1$$

Steps for Completing the Square

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

$$x^{2}+6x+9=2x\left(x+7\right)$$

Use the distributive property to multiply 2x by x+7.

$$x^{2}+6x+9=2x^{2}+14x$$

Subtract 2x^{2} from both sides.

$$x^{2}+6x+9-2x^{2}=14x$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+6x+9=14x$$

Subtract 14x from both sides.

$$-x^{2}+6x+9-14x=0$$

Combine 6x and -14x to get -8x.

$$-x^{2}-8x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}-8x=-9$$

Divide both sides by -1.

$$\frac{-x^{2}-8x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{8}{-1}\right)x=-\frac{9}{-1}$$

Divide -8 by -1.

$$x^{2}+8x=-\frac{9}{-1}$$

Divide -9 by -1.

$$x^{2}+8x=9$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=9+4^{2}$$

Square 4.

$$x^{2}+8x+16=9+16$$

Add 9 to 16.

$$x^{2}+8x+16=25$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+4=5 x+4=-5$$

Subtract 4 from both sides of the equation.

$$x=1 x=-9$$