Solve (xsqrt{2})^2-5x(x-4)=(2sqrt{6})^2-(x-4)^2

$$( x \sqrt { 2 } ) ^ { 2 } - 5 x ( x - 4 ) = ( 2 \sqrt { 6 } ) ^ { 2 } - ( x - 4 ) ^ { 2 }$$

Solve for x

$$x=\sqrt{5}+3\approx 5.236067977$$

$$x=3-\sqrt{5}\approx 0.763932023$$

Steps Using the Quadratic Formula

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

$$x^{2}\left(\sqrt{2}\right)^{2}-5x\left(x-4\right)=\left(2\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}\times 2-5x\left(x-4\right)=\left(2\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

Expand \left(2\sqrt{6}\right)^{2}.

$$x^{2}\times 2-5x\left(x-4\right)=2^{2}\left(\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}\times 2-5x\left(x-4\right)=4\left(\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

The square of \sqrt{6} is 6.

$$x^{2}\times 2-5x\left(x-4\right)=4\times 6-\left(x-4\right)^{2}$$

Multiply 4 and 6 to get 24.

$$x^{2}\times 2-5x\left(x-4\right)=24-\left(x-4\right)^{2}$$

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

$$x^{2}\times 2-5x\left(x-4\right)=24-\left(x^{2}-8x+16\right)$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$x^{2}\times 2-5x\left(x-4\right)=24-x^{2}+8x-16$$

Subtract 16 from 24 to get 8.

$$x^{2}\times 2-5x\left(x-4\right)=8-x^{2}+8x$$

Subtract 8 from both sides.

$$x^{2}\times 2-5x\left(x-4\right)-8=-x^{2}+8x$$

Add x^{2} to both sides.

$$x^{2}\times 2-5x\left(x-4\right)-8+x^{2}=8x$$

Subtract 8x from both sides.

$$x^{2}\times 2-5x\left(x-4\right)-8+x^{2}-8x=0$$

Use the distributive property to multiply -5x by x-4.

$$x^{2}\times 2-5x^{2}+20x-8+x^{2}-8x=0$$

Combine x^{2}\times 2 and -5x^{2} to get -3x^{2}.

$$-3x^{2}+20x-8+x^{2}-8x=0$$

Combine -3x^{2} and x^{2} to get -2x^{2}.

$$-2x^{2}+20x-8-8x=0$$

Combine 20x and -8x to get 12x.

$$-2x^{2}+12x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-8\right)}}{2\left(-2\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-2\right)\left(-8\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-12±\sqrt{144+8\left(-8\right)}}{2\left(-2\right)}$$

Multiply 8 times -8.

$$x=\frac{-12±\sqrt{144-64}}{2\left(-2\right)}$$

Add 144 to -64.

$$x=\frac{-12±\sqrt{80}}{2\left(-2\right)}$$

Take the square root of 80.

$$x=\frac{-12±4\sqrt{5}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-12±4\sqrt{5}}{-4}$$

Now solve the equation x=\frac{-12±4\sqrt{5}}{-4} when ± is plus. Add -12 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-12}{-4}$$

Divide -12+4\sqrt{5} by -4.

$$x=3-\sqrt{5}$$

Now solve the equation x=\frac{-12±4\sqrt{5}}{-4} when ± is minus. Subtract 4\sqrt{5} from -12.

$$x=\frac{-4\sqrt{5}-12}{-4}$$

Divide -12-4\sqrt{5} by -4.

$$x=\sqrt{5}+3$$

The equation is now solved.

$$x=3-\sqrt{5} x=\sqrt{5}+3$$

Steps for Completing the Square

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

$$x^{2}\left(\sqrt{2}\right)^{2}-5x\left(x-4\right)=\left(2\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}\times 2-5x\left(x-4\right)=\left(2\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

Expand \left(2\sqrt{6}\right)^{2}.

$$x^{2}\times 2-5x\left(x-4\right)=2^{2}\left(\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}\times 2-5x\left(x-4\right)=4\left(\sqrt{6}\right)^{2}-\left(x-4\right)^{2}$$

The square of \sqrt{6} is 6.

$$x^{2}\times 2-5x\left(x-4\right)=4\times 6-\left(x-4\right)^{2}$$

Multiply 4 and 6 to get 24.

$$x^{2}\times 2-5x\left(x-4\right)=24-\left(x-4\right)^{2}$$

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

$$x^{2}\times 2-5x\left(x-4\right)=24-\left(x^{2}-8x+16\right)$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$x^{2}\times 2-5x\left(x-4\right)=24-x^{2}+8x-16$$

Subtract 16 from 24 to get 8.

$$x^{2}\times 2-5x\left(x-4\right)=8-x^{2}+8x$$

Add x^{2} to both sides.

$$x^{2}\times 2-5x\left(x-4\right)+x^{2}=8+8x$$

Subtract 8x from both sides.

$$x^{2}\times 2-5x\left(x-4\right)+x^{2}-8x=8$$

Use the distributive property to multiply -5x by x-4.

$$x^{2}\times 2-5x^{2}+20x+x^{2}-8x=8$$

Combine x^{2}\times 2 and -5x^{2} to get -3x^{2}.

$$-3x^{2}+20x+x^{2}-8x=8$$

Combine -3x^{2} and x^{2} to get -2x^{2}.

$$-2x^{2}+20x-8x=8$$

Combine 20x and -8x to get 12x.

$$-2x^{2}+12x=8$$

Divide both sides by -2.

$$\frac{-2x^{2}+12x}{-2}=\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{12}{-2}x=\frac{8}{-2}$$

Divide 12 by -2.

$$x^{2}-6x=\frac{8}{-2}$$

Divide 8 by -2.

$$x^{2}-6x=-4$$

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.

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

Square -3.

$$x^{2}-6x+9=-4+9$$

Add -4 to 9.

$$x^{2}-6x+9=5$$

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}.

$$\left(x-3\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-3=\sqrt{5} x-3=-\sqrt{5}$$

Add 3 to both sides of the equation.

$$x=\sqrt{5}+3 x=3-\sqrt{5}$$