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

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

Solve for x

$$x=1$$

$$x = \frac{13}{5} = 2\frac{3}{5} = 2.6$$

Steps Using Factoring By Grouping

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

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

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

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

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

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

Combine 2x and 4x to get 6x.

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

Add 1 and 4 to get 5.

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

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

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

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

$$2x^{2}+6x+5-2x^{2}+12x-18=5x^{2}$$

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

$$6x+5+12x-18=5x^{2}$$

Combine 6x and 12x to get 18x.

$$18x+5-18=5x^{2}$$

Subtract 18 from 5 to get -13.

$$18x-13=5x^{2}$$

Subtract 5x^{2} from both sides.

$$18x-13-5x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5x^{2}+18x-13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx-13. To find a and b, set up a system to be solved.

$$a+b=18 ab=-5\left(-13\right)=65$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 65.

$$1,65 5,13$$

Calculate the sum for each pair.

$$1+65=66 5+13=18$$

The solution is the pair that gives sum 18.

$$a=13 b=5$$

Rewrite -5x^{2}+18x-13 as \left(-5x^{2}+13x\right)+\left(5x-13\right).

$$\left(-5x^{2}+13x\right)+\left(5x-13\right)$$

Factor out -x in -5x^{2}+13x.

$$-x\left(5x-13\right)+5x-13$$

Factor out common term 5x-13 by using distributive property.

$$\left(5x-13\right)\left(-x+1\right)$$

To find equation solutions, solve 5x-13=0 and -x+1=0.

$$x=\frac{13}{5} x=1$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Combine 2x and 4x to get 6x.

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

Add 1 and 4 to get 5.

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

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

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

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

$$2x^{2}+6x+5-2x^{2}+12x-18=5x^{2}$$

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

$$6x+5+12x-18=5x^{2}$$

Combine 6x and 12x to get 18x.

$$18x+5-18=5x^{2}$$

Subtract 18 from 5 to get -13.

$$18x-13=5x^{2}$$

Subtract 5x^{2} from both sides.

$$18x-13-5x^{2}=0$$

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.

$$-5x^{2}+18x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 18 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-5\right)\left(-13\right)}}{2\left(-5\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-5\right)\left(-13\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-18±\sqrt{324+20\left(-13\right)}}{2\left(-5\right)}$$

Multiply 20 times -13.

$$x=\frac{-18±\sqrt{324-260}}{2\left(-5\right)}$$

Add 324 to -260.

$$x=\frac{-18±\sqrt{64}}{2\left(-5\right)}$$

Take the square root of 64.

$$x=\frac{-18±8}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-18±8}{-10}$$

Now solve the equation x=\frac{-18±8}{-10} when ± is plus. Add -18 to 8.

$$x=-\frac{10}{-10}$$

Divide -10 by -10.

$$x=1$$

Now solve the equation x=\frac{-18±8}{-10} when ± is minus. Subtract 8 from -18.

$$x=-\frac{26}{-10}$$

Reduce the fraction \frac{-26}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{5}$$

The equation is now solved.

$$x=1 x=\frac{13}{5}$$

Steps for Completing the Square

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

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

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

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

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

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

Combine 2x and 4x to get 6x.

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

Add 1 and 4 to get 5.

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

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

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

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

$$2x^{2}+6x+5-2x^{2}+12x-18=5x^{2}$$

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

$$6x+5+12x-18=5x^{2}$$

Combine 6x and 12x to get 18x.

$$18x+5-18=5x^{2}$$

Subtract 18 from 5 to get -13.

$$18x-13=5x^{2}$$

Subtract 5x^{2} from both sides.

$$18x-13-5x^{2}=0$$

Add 13 to both sides. Anything plus zero gives itself.

$$18x-5x^{2}=13$$

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.

$$-5x^{2}+18x=13$$

Divide both sides by -5.

$$\frac{-5x^{2}+18x}{-5}=\frac{13}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{18}{-5}x=\frac{13}{-5}$$

Divide 18 by -5.

$$x^{2}-\frac{18}{5}x=\frac{13}{-5}$$

Divide 13 by -5.

$$x^{2}-\frac{18}{5}x=-\frac{13}{5}$$

Divide -\frac{18}{5}, the coefficient of the x term, by 2 to get -\frac{9}{5}. Then add the square of -\frac{9}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=-\frac{13}{5}+\left(-\frac{9}{5}\right)^{2}$$

Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=-\frac{13}{5}+\frac{81}{25}$$

Add -\frac{13}{5} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{16}{25}$$

Factor x^{2}-\frac{18}{5}x+\frac{81}{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}.

$$\left(x-\frac{9}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$x-\frac{9}{5}=\frac{4}{5} x-\frac{9}{5}=-\frac{4}{5}$$

Add \frac{9}{5} to both sides of the equation.

$$x=\frac{13}{5} x=1$$