Solve {left(2sqrt{3}right)}^2=1/2xleft(6+xright)

$${ \left(2 \sqrt{ 3 } \right) }^{ 2 } = \frac{ 1 }{ 2 } x \left( 6+x \right)$$

Solve for x (complex solution)

$$x=\sqrt{33}-3\approx 2.744562647$$

$$x=-\left(\sqrt{33}+3\right)\approx -8.744562647$$

Steps Using the Quadratic Formula

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

The square of \sqrt{3} is 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

Multiply 4 and 3 to get 12.

$$12=\frac{1}{2}x\left(6+x\right)$$

Use the distributive property to multiply \frac{1}{2}x by 6+x.

$$12=3x+\frac{1}{2}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$3x+\frac{1}{2}x^{2}=12$$

Subtract 12 from both sides.

$$3x+\frac{1}{2}x^{2}-12=0$$

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.

$$\frac{1}{2}x^{2}+3x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, 3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-3±\sqrt{9-2\left(-12\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -12.

$$x=\frac{-3±\sqrt{9+24}}{2\times \frac{1}{2}}$$

Add 9 to 24.

$$x=\frac{-3±\sqrt{33}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{-3±\sqrt{33}}{1}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{1} when ± is plus. Add -3 to \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{1}$$

Divide -3+\sqrt{33} by 1.

$$x=\sqrt{33}-3$$

Now solve the equation x=\frac{-3±\sqrt{33}}{1} when ± is minus. Subtract \sqrt{33} from -3.

$$x=\frac{-\sqrt{33}-3}{1}$$

Divide -3-\sqrt{33} by 1.

$$x=-\sqrt{33}-3$$

The equation is now solved.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Steps for Completing the Square

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

The square of \sqrt{3} is 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

Multiply 4 and 3 to get 12.

$$12=\frac{1}{2}x\left(6+x\right)$$

Use the distributive property to multiply \frac{1}{2}x by 6+x.

$$12=3x+\frac{1}{2}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$3x+\frac{1}{2}x^{2}=12$$

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.

$$\frac{1}{2}x^{2}+3x=12$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+3x}{\frac{1}{2}}=\frac{12}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{3}{\frac{1}{2}}x=\frac{12}{\frac{1}{2}}$$

Divide 3 by \frac{1}{2} by multiplying 3 by the reciprocal of \frac{1}{2}.

$$x^{2}+6x=\frac{12}{\frac{1}{2}}$$

Divide 12 by \frac{1}{2} by multiplying 12 by the reciprocal of \frac{1}{2}.

$$x^{2}+6x=24$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=24+3^{2}$$

Square 3.

$$x^{2}+6x+9=24+9$$

Add 24 to 9.

$$x^{2}+6x+9=33$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=33$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{33}$$

Simplify.

$$x+3=\sqrt{33} x+3=-\sqrt{33}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Solve for x

$$x=\sqrt{33}-3\approx 2.744562647$$

$$x=-\sqrt{33}-3\approx -8.744562647$$

Steps Using the Quadratic Formula

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

The square of \sqrt{3} is 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

Multiply 4 and 3 to get 12.

$$12=\frac{1}{2}x\left(6+x\right)$$

Use the distributive property to multiply \frac{1}{2}x by 6+x.

$$12=3x+\frac{1}{2}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$3x+\frac{1}{2}x^{2}=12$$

Subtract 12 from both sides.

$$3x+\frac{1}{2}x^{2}-12=0$$

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.

$$\frac{1}{2}x^{2}+3x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, 3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times \frac{1}{2}\left(-12\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-3±\sqrt{9-2\left(-12\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -12.

$$x=\frac{-3±\sqrt{9+24}}{2\times \frac{1}{2}}$$

Add 9 to 24.

$$x=\frac{-3±\sqrt{33}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{-3±\sqrt{33}}{1}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{1} when ± is plus. Add -3 to \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{1}$$

Divide -3+\sqrt{33} by 1.

$$x=\sqrt{33}-3$$

Now solve the equation x=\frac{-3±\sqrt{33}}{1} when ± is minus. Subtract \sqrt{33} from -3.

$$x=\frac{-\sqrt{33}-3}{1}$$

Divide -3-\sqrt{33} by 1.

$$x=-\sqrt{33}-3$$

The equation is now solved.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$

Steps for Completing the Square

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}=\frac{1}{2}x\left(6+x\right)$$

The square of \sqrt{3} is 3.

$$4\times 3=\frac{1}{2}x\left(6+x\right)$$

Multiply 4 and 3 to get 12.

$$12=\frac{1}{2}x\left(6+x\right)$$

Use the distributive property to multiply \frac{1}{2}x by 6+x.

$$12=3x+\frac{1}{2}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$3x+\frac{1}{2}x^{2}=12$$

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.

$$\frac{1}{2}x^{2}+3x=12$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+3x}{\frac{1}{2}}=\frac{12}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{3}{\frac{1}{2}}x=\frac{12}{\frac{1}{2}}$$

Divide 3 by \frac{1}{2} by multiplying 3 by the reciprocal of \frac{1}{2}.

$$x^{2}+6x=\frac{12}{\frac{1}{2}}$$

Divide 12 by \frac{1}{2} by multiplying 12 by the reciprocal of \frac{1}{2}.

$$x^{2}+6x=24$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=24+3^{2}$$

Square 3.

$$x^{2}+6x+9=24+9$$

Add 24 to 9.

$$x^{2}+6x+9=33$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=33$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{33}$$

Simplify.

$$x+3=\sqrt{33} x+3=-\sqrt{33}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{33}-3 x=-\sqrt{33}-3$$