Solve (4sqrt{3}+frac{xsqrt{3}}{2})^2+x^2/4=156

$$( 4 \sqrt { 3 } + \frac { x \sqrt { 3 } } { 2 } ) ^ { 2 } + \frac { x ^ { 2 } } { 4 } = 156$$

Solve for x

$$x=-18$$

$$x=6$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$4\left(4\sqrt{3}+\frac{x\sqrt{3}}{2}\right)^{2}+x^{2}=624$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4\sqrt{3}+\frac{x\sqrt{3}}{2}\right)^{2}.

$$4\left(16\left(\sqrt{3}\right)^{2}+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

The square of \sqrt{3} is 3.

$$4\left(16\times 3+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

Multiply 16 and 3 to get 48.

$$4\left(48+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

Cancel out 2, the greatest common factor in 8 and 2.

$$4\left(48+4x\sqrt{3}\sqrt{3}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

To raise \frac{x\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4\left(48+4x\sqrt{3}\sqrt{3}+\frac{\left(x\sqrt{3}\right)^{2}}{2^{2}}\right)+x^{2}=624$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 48 times \frac{2^{2}}{2^{2}}.

$$4\left(\frac{48\times 2^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}+\frac{\left(x\sqrt{3}\right)^{2}}{2^{2}}\right)+x^{2}=624$$

Since \frac{48\times 2^{2}}{2^{2}} and \frac{\left(x\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$4\left(\frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}\right)+x^{2}=624$$

Use the distributive property to multiply 4 by \frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}.

$$4\times \frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Calculate 2 to the power of 2 and get 4.

$$4\times \frac{48\times 4+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Multiply 48 and 4 to get 192.

$$4\times \frac{192+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

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

$$4\times \frac{192+x^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

The square of \sqrt{3} is 3.

$$4\times \frac{192+x^{2}\times 3}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Calculate 2 to the power of 2 and get 4.

$$4\times \frac{192+x^{2}\times 3}{4}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Express 4\times \frac{192+x^{2}\times 3}{4} as a single fraction.

$$\frac{4\left(192+x^{2}\times 3\right)}{4}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Cancel out 4 and 4.

$$192+x^{2}\times 3+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

The square of \sqrt{3} is 3.

$$192+x^{2}\times 3+16\times 3x+x^{2}=624$$

Multiply 16 and 3 to get 48.

$$192+x^{2}\times 3+48x+x^{2}=624$$

Combine x^{2}\times 3 and x^{2} to get 4x^{2}.

$$192+4x^{2}+48x=624$$

Subtract 624 from both sides.

$$192+4x^{2}+48x-624=0$$

Subtract 624 from 192 to get -432.

$$-432+4x^{2}+48x=0$$

Divide both sides by 4.

$$-108+x^{2}+12x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+12x-108=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-108. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\left(-108\right)=-108$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -108.

$$-1,108 -2,54 -3,36 -4,27 -6,18 -9,12$$

Calculate the sum for each pair.

$$-1+108=107 -2+54=52 -3+36=33 -4+27=23 -6+18=12 -9+12=3$$

The solution is the pair that gives sum 12.

$$a=-6 b=18$$

Rewrite x^{2}+12x-108 as \left(x^{2}-6x\right)+\left(18x-108\right).

$$\left(x^{2}-6x\right)+\left(18x-108\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-6\right)+18\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+18\right)$$

To find equation solutions, solve x-6=0 and x+18=0.

$$x=6 x=-18$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$4\left(4\sqrt{3}+\frac{x\sqrt{3}}{2}\right)^{2}+x^{2}=624$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4\sqrt{3}+\frac{x\sqrt{3}}{2}\right)^{2}.

$$4\left(16\left(\sqrt{3}\right)^{2}+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

The square of \sqrt{3} is 3.

$$4\left(16\times 3+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

Multiply 16 and 3 to get 48.

$$4\left(48+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

Cancel out 2, the greatest common factor in 8 and 2.

$$4\left(48+4x\sqrt{3}\sqrt{3}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

To raise \frac{x\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4\left(48+4x\sqrt{3}\sqrt{3}+\frac{\left(x\sqrt{3}\right)^{2}}{2^{2}}\right)+x^{2}=624$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 48 times \frac{2^{2}}{2^{2}}.

$$4\left(\frac{48\times 2^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}+\frac{\left(x\sqrt{3}\right)^{2}}{2^{2}}\right)+x^{2}=624$$

Since \frac{48\times 2^{2}}{2^{2}} and \frac{\left(x\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$4\left(\frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}\right)+x^{2}=624$$

Use the distributive property to multiply 4 by \frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}.

$$4\times \frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Calculate 2 to the power of 2 and get 4.

$$4\times \frac{48\times 4+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Multiply 48 and 4 to get 192.

$$4\times \frac{192+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

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

$$4\times \frac{192+x^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

The square of \sqrt{3} is 3.

$$4\times \frac{192+x^{2}\times 3}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Calculate 2 to the power of 2 and get 4.

$$4\times \frac{192+x^{2}\times 3}{4}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Express 4\times \frac{192+x^{2}\times 3}{4} as a single fraction.

$$\frac{4\left(192+x^{2}\times 3\right)}{4}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Cancel out 4 and 4.

$$192+x^{2}\times 3+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

The square of \sqrt{3} is 3.

$$192+x^{2}\times 3+16\times 3x+x^{2}=624$$

Multiply 16 and 3 to get 48.

$$192+x^{2}\times 3+48x+x^{2}=624$$

Combine x^{2}\times 3 and x^{2} to get 4x^{2}.

$$192+4x^{2}+48x=624$$

Subtract 624 from both sides.

$$192+4x^{2}+48x-624=0$$

Subtract 624 from 192 to get -432.

$$-432+4x^{2}+48x=0$$

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.

$$4x^{2}+48x-432=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 48 for b, and -432 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-48±\sqrt{48^{2}-4\times 4\left(-432\right)}}{2\times 4}$$

Square 48.

$$x=\frac{-48±\sqrt{2304-4\times 4\left(-432\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-48±\sqrt{2304-16\left(-432\right)}}{2\times 4}$$

Multiply -16 times -432.

$$x=\frac{-48±\sqrt{2304+6912}}{2\times 4}$$

Add 2304 to 6912.

$$x=\frac{-48±\sqrt{9216}}{2\times 4}$$

Take the square root of 9216.

$$x=\frac{-48±96}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-48±96}{8}$$

Now solve the equation x=\frac{-48±96}{8} when ± is plus. Add -48 to 96.

$$x=\frac{48}{8}$$

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{-48±96}{8} when ± is minus. Subtract 96 from -48.

$$x=-\frac{144}{8}$$

Divide -144 by 8.

$$x=-18$$

The equation is now solved.

$$x=6 x=-18$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$4\left(4\sqrt{3}+\frac{x\sqrt{3}}{2}\right)^{2}+x^{2}=624$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4\sqrt{3}+\frac{x\sqrt{3}}{2}\right)^{2}.

$$4\left(16\left(\sqrt{3}\right)^{2}+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

The square of \sqrt{3} is 3.

$$4\left(16\times 3+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

Multiply 16 and 3 to get 48.

$$4\left(48+8\sqrt{3}\times \frac{x\sqrt{3}}{2}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

Cancel out 2, the greatest common factor in 8 and 2.

$$4\left(48+4x\sqrt{3}\sqrt{3}+\left(\frac{x\sqrt{3}}{2}\right)^{2}\right)+x^{2}=624$$

To raise \frac{x\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4\left(48+4x\sqrt{3}\sqrt{3}+\frac{\left(x\sqrt{3}\right)^{2}}{2^{2}}\right)+x^{2}=624$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 48 times \frac{2^{2}}{2^{2}}.

$$4\left(\frac{48\times 2^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}+\frac{\left(x\sqrt{3}\right)^{2}}{2^{2}}\right)+x^{2}=624$$

Since \frac{48\times 2^{2}}{2^{2}} and \frac{\left(x\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$4\left(\frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}\right)+x^{2}=624$$

Use the distributive property to multiply 4 by \frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+4x\sqrt{3}\sqrt{3}.

$$4\times \frac{48\times 2^{2}+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Calculate 2 to the power of 2 and get 4.

$$4\times \frac{48\times 4+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Multiply 48 and 4 to get 192.

$$4\times \frac{192+\left(x\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

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

$$4\times \frac{192+x^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

The square of \sqrt{3} is 3.

$$4\times \frac{192+x^{2}\times 3}{2^{2}}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Calculate 2 to the power of 2 and get 4.

$$4\times \frac{192+x^{2}\times 3}{4}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Express 4\times \frac{192+x^{2}\times 3}{4} as a single fraction.

$$\frac{4\left(192+x^{2}\times 3\right)}{4}+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

Cancel out 4 and 4.

$$192+x^{2}\times 3+16\left(\sqrt{3}\right)^{2}x+x^{2}=624$$

The square of \sqrt{3} is 3.

$$192+x^{2}\times 3+16\times 3x+x^{2}=624$$

Multiply 16 and 3 to get 48.

$$192+x^{2}\times 3+48x+x^{2}=624$$

Combine x^{2}\times 3 and x^{2} to get 4x^{2}.

$$192+4x^{2}+48x=624$$

Subtract 192 from both sides.

$$4x^{2}+48x=624-192$$

Subtract 192 from 624 to get 432.

$$4x^{2}+48x=432$$

Divide both sides by 4.

$$\frac{4x^{2}+48x}{4}=\frac{432}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{48}{4}x=\frac{432}{4}$$

Divide 48 by 4.

$$x^{2}+12x=\frac{432}{4}$$

Divide 432 by 4.

$$x^{2}+12x=108$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=108+6^{2}$$

Square 6.

$$x^{2}+12x+36=108+36$$

Add 108 to 36.

$$x^{2}+12x+36=144$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{144}$$

Simplify.

$$x+6=12 x+6=-12$$

Subtract 6 from both sides of the equation.

$$x=6 x=-18$$