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

$$2 \cdot ( x + 7 ) = ( x + 2 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=\sqrt{11}-1\approx 2.31662479$$

$$x=-\left(\sqrt{11}+1\right)\approx -4.31662479$$

Steps Using the Quadratic Formula

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

$$2x+14=\left(x+2\right)^{2}$$

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

$$2x+14=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$2x+14-x^{2}=4x+4$$

Subtract 4x from both sides.

$$2x+14-x^{2}-4x=4$$

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

$$-2x+14-x^{2}=4$$

Subtract 4 from both sides.

$$-2x+14-x^{2}-4=0$$

Subtract 4 from 14 to get 10.

$$-2x+10-x^{2}=0$$

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.

$$-x^{2}-2x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 10}}{2\left(-1\right)}$$

Multiply 4 times 10.

$$x=\frac{-\left(-2\right)±\sqrt{4+40}}{2\left(-1\right)}$$

Add 4 to 40.

$$x=\frac{-\left(-2\right)±\sqrt{44}}{2\left(-1\right)}$$

Take the square root of 44.

$$x=\frac{-\left(-2\right)±2\sqrt{11}}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{11}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{11}}{-2}$$

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

$$x=\frac{2\sqrt{11}+2}{-2}$$

Divide 2+2\sqrt{11} by -2.

$$x=-\left(\sqrt{11}+1\right)$$

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

$$x=\frac{2-2\sqrt{11}}{-2}$$

Divide 2-2\sqrt{11} by -2.

$$x=\sqrt{11}-1$$

The equation is now solved.

$$x=-\left(\sqrt{11}+1\right) x=\sqrt{11}-1$$

Steps for Completing the Square

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

$$2x+14=\left(x+2\right)^{2}$$

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

$$2x+14=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$2x+14-x^{2}=4x+4$$

Subtract 4x from both sides.

$$2x+14-x^{2}-4x=4$$

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

$$-2x+14-x^{2}=4$$

Subtract 14 from both sides.

$$-2x-x^{2}=4-14$$

Subtract 14 from 4 to get -10.

$$-2x-x^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-2x=-10$$

Divide both sides by -1.

$$\frac{-x^{2}-2x}{-1}=-\frac{10}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{10}{-1}$$

Divide -2 by -1.

$$x^{2}+2x=-\frac{10}{-1}$$

Divide -10 by -1.

$$x^{2}+2x=10$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=10+1^{2}$$

Square 1.

$$x^{2}+2x+1=10+1$$

Add 10 to 1.

$$x^{2}+2x+1=11$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{11}$$

Simplify.

$$x+1=\sqrt{11} x+1=-\sqrt{11}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{11}-1 x=-\sqrt{11}-1$$

Solve for x

$$x=\sqrt{11}-1\approx 2.31662479$$

$$x=-\sqrt{11}-1\approx -4.31662479$$

Steps Using the Quadratic Formula

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

$$2x+14=\left(x+2\right)^{2}$$

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

$$2x+14=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$2x+14-x^{2}=4x+4$$

Subtract 4x from both sides.

$$2x+14-x^{2}-4x=4$$

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

$$-2x+14-x^{2}=4$$

Subtract 4 from both sides.

$$-2x+14-x^{2}-4=0$$

Subtract 4 from 14 to get 10.

$$-2x+10-x^{2}=0$$

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.

$$-x^{2}-2x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 10}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 10}}{2\left(-1\right)}$$

Multiply 4 times 10.

$$x=\frac{-\left(-2\right)±\sqrt{4+40}}{2\left(-1\right)}$$

Add 4 to 40.

$$x=\frac{-\left(-2\right)±\sqrt{44}}{2\left(-1\right)}$$

Take the square root of 44.

$$x=\frac{-\left(-2\right)±2\sqrt{11}}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{11}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{11}}{-2}$$

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

$$x=\frac{2\sqrt{11}+2}{-2}$$

Divide 2+2\sqrt{11} by -2.

$$x=-\left(\sqrt{11}+1\right)$$

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

$$x=\frac{2-2\sqrt{11}}{-2}$$

Divide 2-2\sqrt{11} by -2.

$$x=\sqrt{11}-1$$

The equation is now solved.

$$x=-\left(\sqrt{11}+1\right) x=\sqrt{11}-1$$

Steps for Completing the Square

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

$$2x+14=\left(x+2\right)^{2}$$

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

$$2x+14=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$2x+14-x^{2}=4x+4$$

Subtract 4x from both sides.

$$2x+14-x^{2}-4x=4$$

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

$$-2x+14-x^{2}=4$$

Subtract 14 from both sides.

$$-2x-x^{2}=4-14$$

Subtract 14 from 4 to get -10.

$$-2x-x^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-2x=-10$$

Divide both sides by -1.

$$\frac{-x^{2}-2x}{-1}=-\frac{10}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{10}{-1}$$

Divide -2 by -1.

$$x^{2}+2x=-\frac{10}{-1}$$

Divide -10 by -1.

$$x^{2}+2x=10$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=10+1^{2}$$

Square 1.

$$x^{2}+2x+1=10+1$$

Add 10 to 1.

$$x^{2}+2x+1=11$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{11}$$

Simplify.

$$x+1=\sqrt{11} x+1=-\sqrt{11}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{11}-1 x=-\sqrt{11}-1$$