Solve x^2+(6-3x)^2+4x+16(6-3x)+28=0

$$x ^ { 2 } + ( 6 - 3 x ) ^ { 2 } + 4 x + 16 ( 6 - 3 x ) + 28 = 0$$

Solve for x

$$x=4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-3x\right)^{2}.

$$x^{2}+36-36x+9x^{2}+4x+16\left(6-3x\right)+28=0$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}+36-36x+4x+16\left(6-3x\right)+28=0$$

Combine -36x and 4x to get -32x.

$$10x^{2}+36-32x+16\left(6-3x\right)+28=0$$

Use the distributive property to multiply 16 by 6-3x.

$$10x^{2}+36-32x+96-48x+28=0$$

Add 36 and 96 to get 132.

$$10x^{2}+132-32x-48x+28=0$$

Combine -32x and -48x to get -80x.

$$10x^{2}+132-80x+28=0$$

Add 132 and 28 to get 160.

$$10x^{2}+160-80x=0$$

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.

$$10x^{2}-80x+160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -80 for b, and 160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 10\times 160}}{2\times 10}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 10\times 160}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-80\right)±\sqrt{6400-40\times 160}}{2\times 10}$$

Multiply -40 times 160.

$$x=\frac{-\left(-80\right)±\sqrt{6400-6400}}{2\times 10}$$

Add 6400 to -6400.

$$x=\frac{-\left(-80\right)±\sqrt{0}}{2\times 10}$$

Take the square root of 0.

$$x=-\frac{-80}{2\times 10}$$

The opposite of -80 is 80.

$$x=\frac{80}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{80}{20}$$

Divide 80 by 20.

$$x=4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-3x\right)^{2}.

$$x^{2}+36-36x+9x^{2}+4x+16\left(6-3x\right)+28=0$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}+36-36x+4x+16\left(6-3x\right)+28=0$$

Combine -36x and 4x to get -32x.

$$10x^{2}+36-32x+16\left(6-3x\right)+28=0$$

Use the distributive property to multiply 16 by 6-3x.

$$10x^{2}+36-32x+96-48x+28=0$$

Add 36 and 96 to get 132.

$$10x^{2}+132-32x-48x+28=0$$

Combine -32x and -48x to get -80x.

$$10x^{2}+132-80x+28=0$$

Add 132 and 28 to get 160.

$$10x^{2}+160-80x=0$$

Subtract 160 from both sides. Anything subtracted from zero gives its negation.

$$10x^{2}-80x=-160$$

Divide both sides by 10.

$$\frac{10x^{2}-80x}{10}=-\frac{160}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{80}{10}\right)x=-\frac{160}{10}$$

Divide -80 by 10.

$$x^{2}-8x=-\frac{160}{10}$$

Divide -160 by 10.

$$x^{2}-8x=-16$$

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.

$$x^{2}-8x+\left(-4\right)^{2}=-16+\left(-4\right)^{2}$$

Square -4.

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

Add -16 to 16.

$$x^{2}-8x+16=0$$

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}.

$$\left(x-4\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-4=0 x-4=0$$

Add 4 to both sides of the equation.

$$x=4 x=4$$

The equation is now solved. Solutions are the same.

$$x=4$$