Solve 4^2+(2-3x)*(2+3x)=(x-4)^2

$$4 ^ { 2 } + ( 2 - 3 x ) \cdot ( 2 + 3 x ) = ( x - 4 ) ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{14} + 2}{5} \approx 1.148331477$$

$$x=\frac{2-\sqrt{14}}{5}\approx -0.348331477$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$16+\left(2-3x\right)\left(2+3x\right)=\left(x-4\right)^{2}$$

Consider \left(2-3x\right)\left(2+3x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$16+4-\left(3x\right)^{2}=\left(x-4\right)^{2}$$

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

$$16+4-3^{2}x^{2}=\left(x-4\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16+4-9x^{2}=\left(x-4\right)^{2}$$

Add 16 and 4 to get 20.

$$20-9x^{2}=\left(x-4\right)^{2}$$

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

$$20-9x^{2}=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$20-9x^{2}-x^{2}=-8x+16$$

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

$$20-10x^{2}=-8x+16$$

Add 8x to both sides.

$$20-10x^{2}+8x=16$$

Subtract 16 from both sides.

$$20-10x^{2}+8x-16=0$$

Subtract 16 from 20 to get 4.

$$4-10x^{2}+8x=0$$

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.

$$-10x^{2}+8x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 8 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-10\right)\times 4}}{2\left(-10\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-10\right)\times 4}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-8±\sqrt{64+40\times 4}}{2\left(-10\right)}$$

Multiply 40 times 4.

$$x=\frac{-8±\sqrt{64+160}}{2\left(-10\right)}$$

Add 64 to 160.

$$x=\frac{-8±\sqrt{224}}{2\left(-10\right)}$$

Take the square root of 224.

$$x=\frac{-8±4\sqrt{14}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-8±4\sqrt{14}}{-20}$$

Now solve the equation x=\frac{-8±4\sqrt{14}}{-20} when ± is plus. Add -8 to 4\sqrt{14}.

$$x=\frac{4\sqrt{14}-8}{-20}$$

Divide -8+4\sqrt{14} by -20.

$$x=\frac{2-\sqrt{14}}{5}$$

Now solve the equation x=\frac{-8±4\sqrt{14}}{-20} when ± is minus. Subtract 4\sqrt{14} from -8.

$$x=\frac{-4\sqrt{14}-8}{-20}$$

Divide -8-4\sqrt{14} by -20.

$$x=\frac{\sqrt{14}+2}{5}$$

The equation is now solved.

$$x=\frac{2-\sqrt{14}}{5} x=\frac{\sqrt{14}+2}{5}$$

Steps for Completing the Square

Calculate 4 to the power of 2 and get 16.

$$16+\left(2-3x\right)\left(2+3x\right)=\left(x-4\right)^{2}$$

Consider \left(2-3x\right)\left(2+3x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$16+4-\left(3x\right)^{2}=\left(x-4\right)^{2}$$

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

$$16+4-3^{2}x^{2}=\left(x-4\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16+4-9x^{2}=\left(x-4\right)^{2}$$

Add 16 and 4 to get 20.

$$20-9x^{2}=\left(x-4\right)^{2}$$

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

$$20-9x^{2}=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$20-9x^{2}-x^{2}=-8x+16$$

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

$$20-10x^{2}=-8x+16$$

Add 8x to both sides.

$$20-10x^{2}+8x=16$$

Subtract 20 from both sides.

$$-10x^{2}+8x=16-20$$

Subtract 20 from 16 to get -4.

$$-10x^{2}+8x=-4$$

Divide both sides by -10.

$$\frac{-10x^{2}+8x}{-10}=-\frac{4}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{8}{-10}x=-\frac{4}{-10}$$

Reduce the fraction \frac{8}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{5}x=-\frac{4}{-10}$$

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{5}x=\frac{2}{5}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{2}{5}+\frac{4}{25}$$

Add \frac{2}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{14}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{14}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{\sqrt{14}}{5} x-\frac{2}{5}=-\frac{\sqrt{14}}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{\sqrt{14}+2}{5} x=\frac{2-\sqrt{14}}{5}$$