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

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

Solve for x

$$x=\frac{\sqrt{5}}{2}-2\approx -0.881966011$$

$$x=-\frac{\sqrt{5}}{2}-2\approx -3.118033989$$

Steps Using the Quadratic Formula

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

$$4\left(x^{2}+6x+9\right)-2\left(x-1\right)^{2}+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36-2\left(x-1\right)^{2}+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

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

$$4x^{2}+24x+36-2\left(x^{2}-2x+1\right)+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply -2 by x^{2}-2x+1.

$$4x^{2}+24x+36-2x^{2}+4x-2+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}+24x+36+4x-2+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Combine 24x and 4x to get 28x.

$$2x^{2}+28x+36-2+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Subtract 2 from 36 to get 34.

$$2x^{2}+28x+34+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply x-5 by x+3 and combine like terms.

$$2x^{2}+28x+34+x^{2}-2x-15+2\left(x^{2}-5x+1\right)=x^{2}+10$$

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

$$3x^{2}+28x+34-2x-15+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Combine 28x and -2x to get 26x.

$$3x^{2}+26x+34-15+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Subtract 15 from 34 to get 19.

$$3x^{2}+26x+19+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply 2 by x^{2}-5x+1.

$$3x^{2}+26x+19+2x^{2}-10x+2=x^{2}+10$$

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

$$5x^{2}+26x+19-10x+2=x^{2}+10$$

Combine 26x and -10x to get 16x.

$$5x^{2}+16x+19+2=x^{2}+10$$

Add 19 and 2 to get 21.

$$5x^{2}+16x+21=x^{2}+10$$

Subtract x^{2} from both sides.

$$5x^{2}+16x+21-x^{2}=10$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+16x+21=10$$

Subtract 10 from both sides.

$$4x^{2}+16x+21-10=0$$

Subtract 10 from 21 to get 11.

$$4x^{2}+16x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 16 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 4\times 11}}{2\times 4}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 4\times 11}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-16±\sqrt{256-16\times 11}}{2\times 4}$$

Multiply -16 times 11.

$$x=\frac{-16±\sqrt{256-176}}{2\times 4}$$

Add 256 to -176.

$$x=\frac{-16±\sqrt{80}}{2\times 4}$$

Take the square root of 80.

$$x=\frac{-16±4\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-16±4\sqrt{5}}{8}$$

Now solve the equation x=\frac{-16±4\sqrt{5}}{8} when ± is plus. Add -16 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-16}{8}$$

Divide -16+4\sqrt{5} by 8.

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

Now solve the equation x=\frac{-16±4\sqrt{5}}{8} when ± is minus. Subtract 4\sqrt{5} from -16.

$$x=\frac{-4\sqrt{5}-16}{8}$$

Divide -16-4\sqrt{5} by 8.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$4\left(x^{2}+6x+9\right)-2\left(x-1\right)^{2}+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36-2\left(x-1\right)^{2}+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

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

$$4x^{2}+24x+36-2\left(x^{2}-2x+1\right)+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply -2 by x^{2}-2x+1.

$$4x^{2}+24x+36-2x^{2}+4x-2+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}+24x+36+4x-2+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Combine 24x and 4x to get 28x.

$$2x^{2}+28x+36-2+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Subtract 2 from 36 to get 34.

$$2x^{2}+28x+34+\left(x-5\right)\left(x+3\right)+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply x-5 by x+3 and combine like terms.

$$2x^{2}+28x+34+x^{2}-2x-15+2\left(x^{2}-5x+1\right)=x^{2}+10$$

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

$$3x^{2}+28x+34-2x-15+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Combine 28x and -2x to get 26x.

$$3x^{2}+26x+34-15+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Subtract 15 from 34 to get 19.

$$3x^{2}+26x+19+2\left(x^{2}-5x+1\right)=x^{2}+10$$

Use the distributive property to multiply 2 by x^{2}-5x+1.

$$3x^{2}+26x+19+2x^{2}-10x+2=x^{2}+10$$

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

$$5x^{2}+26x+19-10x+2=x^{2}+10$$

Combine 26x and -10x to get 16x.

$$5x^{2}+16x+19+2=x^{2}+10$$

Add 19 and 2 to get 21.

$$5x^{2}+16x+21=x^{2}+10$$

Subtract x^{2} from both sides.

$$5x^{2}+16x+21-x^{2}=10$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+16x+21=10$$

Subtract 21 from both sides.

$$4x^{2}+16x=10-21$$

Subtract 21 from 10 to get -11.

$$4x^{2}+16x=-11$$

Divide both sides by 4.

$$\frac{4x^{2}+16x}{4}=-\frac{11}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{16}{4}x=-\frac{11}{4}$$

Divide 16 by 4.

$$x^{2}+4x=-\frac{11}{4}$$

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.

$$x^{2}+4x+2^{2}=-\frac{11}{4}+2^{2}$$

Square 2.

$$x^{2}+4x+4=-\frac{11}{4}+4$$

Add -\frac{11}{4} to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 2 from both sides of the equation.

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