Solve (x+3)-10-(x-2)^2=(x-1)^2-(x-5)^2+4+x

$$( x + 3 ) - 10 - ( x - 2 ) ^ { 2 } = ( x - 1 ) ^ { 2 } - ( x - 5 ) ^ { 2 } + 4 + x$$

Solve for x

$$x=\sqrt{13}-2\approx 1.605551275$$

$$x=-\left(\sqrt{13}+2\right)\approx -5.605551275$$

Steps Using the Quadratic Formula

Subtract 10 from 3 to get -7.

$$x-7-\left(x-2\right)^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-\left(x^{2}-4x+4\right)=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-x^{2}+4x-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Combine x and 4x to get 5x.

$$5x-7-x^{2}-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Subtract 4 from -7 to get -11.

$$5x-11-x^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x-5\right)^{2}+4+x$$

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

$$5x-11-x^{2}=x^{2}-2x+1-\left(x^{2}-10x+25\right)+4+x$$

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

$$5x-11-x^{2}=x^{2}-2x+1-x^{2}+10x-25+4+x$$

Combine x^{2} and -x^{2} to get 0.

$$5x-11-x^{2}=-2x+1+10x-25+4+x$$

Combine -2x and 10x to get 8x.

$$5x-11-x^{2}=8x+1-25+4+x$$

Subtract 25 from 1 to get -24.

$$5x-11-x^{2}=8x-24+4+x$$

Add -24 and 4 to get -20.

$$5x-11-x^{2}=8x-20+x$$

Combine 8x and x to get 9x.

$$5x-11-x^{2}=9x-20$$

Subtract 9x from both sides.

$$5x-11-x^{2}-9x=-20$$

Combine 5x and -9x to get -4x.

$$-4x-11-x^{2}=-20$$

Add 20 to both sides.

$$-4x-11-x^{2}+20=0$$

Add -11 and 20 to get 9.

$$-4x+9-x^{2}=0$$

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.

$$-x^{2}-4x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-\left(-4\right)±\sqrt{16+36}}{2\left(-1\right)}$$

Add 16 to 36.

$$x=\frac{-\left(-4\right)±\sqrt{52}}{2\left(-1\right)}$$

Take the square root of 52.

$$x=\frac{-\left(-4\right)±2\sqrt{13}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{13}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{13}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{13}}{-2} when ± is plus. Add 4 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+4}{-2}$$

Divide 4+2\sqrt{13} by -2.

$$x=-\left(\sqrt{13}+2\right)$$

Now solve the equation x=\frac{4±2\sqrt{13}}{-2} when ± is minus. Subtract 2\sqrt{13} from 4.

$$x=\frac{4-2\sqrt{13}}{-2}$$

Divide 4-2\sqrt{13} by -2.

$$x=\sqrt{13}-2$$

The equation is now solved.

$$x=-\left(\sqrt{13}+2\right) x=\sqrt{13}-2$$

Steps for Completing the Square

Subtract 10 from 3 to get -7.

$$x-7-\left(x-2\right)^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-\left(x^{2}-4x+4\right)=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-x^{2}+4x-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Combine x and 4x to get 5x.

$$5x-7-x^{2}-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Subtract 4 from -7 to get -11.

$$5x-11-x^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x-5\right)^{2}+4+x$$

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

$$5x-11-x^{2}=x^{2}-2x+1-\left(x^{2}-10x+25\right)+4+x$$

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

$$5x-11-x^{2}=x^{2}-2x+1-x^{2}+10x-25+4+x$$

Combine x^{2} and -x^{2} to get 0.

$$5x-11-x^{2}=-2x+1+10x-25+4+x$$

Combine -2x and 10x to get 8x.

$$5x-11-x^{2}=8x+1-25+4+x$$

Subtract 25 from 1 to get -24.

$$5x-11-x^{2}=8x-24+4+x$$

Add -24 and 4 to get -20.

$$5x-11-x^{2}=8x-20+x$$

Combine 8x and x to get 9x.

$$5x-11-x^{2}=9x-20$$

Subtract 9x from both sides.

$$5x-11-x^{2}-9x=-20$$

Combine 5x and -9x to get -4x.

$$-4x-11-x^{2}=-20$$

Add 11 to both sides.

$$-4x-x^{2}=-20+11$$

Add -20 and 11 to get -9.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{9}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=-\frac{9}{-1}$$

Divide -9 by -1.

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

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

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

Square 2.

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

Add 9 to 4.

$$x^{2}+4x+4=13$$

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

$$\left(x+2\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{13}$$

Simplify.

$$x+2=\sqrt{13} x+2=-\sqrt{13}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{13}-2 x=-\sqrt{13}-2$$