Solve x^2-(3x-2)^2=5-(x-5)^2

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

Solve for x

$$x = \frac{\sqrt{113} + 1}{7} \approx 1.661449402$$

$$x=\frac{1-\sqrt{113}}{7}\approx -1.375735116$$

Steps Using the Quadratic Formula

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

$$x^{2}-\left(9x^{2}-12x+4\right)=5-\left(x-5\right)^{2}$$

To find the opposite of 9x^{2}-12x+4, find the opposite of each term.

$$x^{2}-9x^{2}+12x-4=5-\left(x-5\right)^{2}$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}+12x-4=5-\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-8x^{2}+12x-4=5-\left(x^{2}-10x+25\right)$$

To find the opposite of x^{2}-10x+25, find the opposite of each term.

$$-8x^{2}+12x-4=5-x^{2}+10x-25$$

Subtract 25 from 5 to get -20.

$$-8x^{2}+12x-4=-20-x^{2}+10x$$

Subtract -20 from both sides.

$$-8x^{2}+12x-4-\left(-20\right)=-x^{2}+10x$$

The opposite of -20 is 20.

$$-8x^{2}+12x-4+20=-x^{2}+10x$$

Add x^{2} to both sides.

$$-8x^{2}+12x-4+20+x^{2}=10x$$

Add -4 and 20 to get 16.

$$-8x^{2}+12x+16+x^{2}=10x$$

Combine -8x^{2} and x^{2} to get -7x^{2}.

$$-7x^{2}+12x+16=10x$$

Subtract 10x from both sides.

$$-7x^{2}+12x+16-10x=0$$

Combine 12x and -10x to get 2x.

$$-7x^{2}+2x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 2 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-7\right)\times 16}}{2\left(-7\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-7\right)\times 16}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-2±\sqrt{4+28\times 16}}{2\left(-7\right)}$$

Multiply 28 times 16.

$$x=\frac{-2±\sqrt{4+448}}{2\left(-7\right)}$$

Add 4 to 448.

$$x=\frac{-2±\sqrt{452}}{2\left(-7\right)}$$

Take the square root of 452.

$$x=\frac{-2±2\sqrt{113}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-2±2\sqrt{113}}{-14}$$

Now solve the equation x=\frac{-2±2\sqrt{113}}{-14} when ± is plus. Add -2 to 2\sqrt{113}.

$$x=\frac{2\sqrt{113}-2}{-14}$$

Divide -2+2\sqrt{113} by -14.

$$x=\frac{1-\sqrt{113}}{7}$$

Now solve the equation x=\frac{-2±2\sqrt{113}}{-14} when ± is minus. Subtract 2\sqrt{113} from -2.

$$x=\frac{-2\sqrt{113}-2}{-14}$$

Divide -2-2\sqrt{113} by -14.

$$x=\frac{\sqrt{113}+1}{7}$$

The equation is now solved.

$$x=\frac{1-\sqrt{113}}{7} x=\frac{\sqrt{113}+1}{7}$$

Steps for Completing the Square

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

$$x^{2}-\left(9x^{2}-12x+4\right)=5-\left(x-5\right)^{2}$$

To find the opposite of 9x^{2}-12x+4, find the opposite of each term.

$$x^{2}-9x^{2}+12x-4=5-\left(x-5\right)^{2}$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}+12x-4=5-\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-8x^{2}+12x-4=5-\left(x^{2}-10x+25\right)$$

To find the opposite of x^{2}-10x+25, find the opposite of each term.

$$-8x^{2}+12x-4=5-x^{2}+10x-25$$

Subtract 25 from 5 to get -20.

$$-8x^{2}+12x-4=-20-x^{2}+10x$$

Add x^{2} to both sides.

$$-8x^{2}+12x-4+x^{2}=-20+10x$$

Combine -8x^{2} and x^{2} to get -7x^{2}.

$$-7x^{2}+12x-4=-20+10x$$

Subtract 10x from both sides.

$$-7x^{2}+12x-4-10x=-20$$

Combine 12x and -10x to get 2x.

$$-7x^{2}+2x-4=-20$$

Add 4 to both sides.

$$-7x^{2}+2x=-20+4$$

Add -20 and 4 to get -16.

$$-7x^{2}+2x=-16$$

Divide both sides by -7.

$$\frac{-7x^{2}+2x}{-7}=-\frac{16}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\frac{2}{-7}x=-\frac{16}{-7}$$

Divide 2 by -7.

$$x^{2}-\frac{2}{7}x=-\frac{16}{-7}$$

Divide -16 by -7.

$$x^{2}-\frac{2}{7}x=\frac{16}{7}$$

Divide -\frac{2}{7}, the coefficient of the x term, by 2 to get -\frac{1}{7}. Then add the square of -\frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{7}x+\left(-\frac{1}{7}\right)^{2}=\frac{16}{7}+\left(-\frac{1}{7}\right)^{2}$$

Square -\frac{1}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{16}{7}+\frac{1}{49}$$

Add \frac{16}{7} to \frac{1}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{113}{49}$$

Factor x^{2}-\frac{2}{7}x+\frac{1}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{7}\right)^{2}=\frac{113}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{7}\right)^{2}}=\sqrt{\frac{113}{49}}$$

Simplify.

$$x-\frac{1}{7}=\frac{\sqrt{113}}{7} x-\frac{1}{7}=-\frac{\sqrt{113}}{7}$$

Add \frac{1}{7} to both sides of the equation.

$$x=\frac{\sqrt{113}+1}{7} x=\frac{1-\sqrt{113}}{7}$$