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

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

Solve for x

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$x=1$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply 2x by 2x+1.

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

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

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

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$4x^{2}+2x-x^{2}+2x-1=6$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+2x+2x-1=6$$

Combine 2x and 2x to get 4x.

$$3x^{2}+4x-1=6$$

Subtract 6 from both sides.

$$3x^{2}+4x-1-6=0$$

Subtract 6 from -1 to get -7.

$$3x^{2}+4x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-7\right)=-21$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -21.

$$-1,21 -3,7$$

Calculate the sum for each pair.

$$-1+21=20 -3+7=4$$

The solution is the pair that gives sum 4.

$$a=-3 b=7$$

Rewrite 3x^{2}+4x-7 as \left(3x^{2}-3x\right)+\left(7x-7\right).

$$\left(3x^{2}-3x\right)+\left(7x-7\right)$$

Factor out 3x in the first and 7 in the second group.

$$3x\left(x-1\right)+7\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x+7\right)$$

To find equation solutions, solve x-1=0 and 3x+7=0.

$$x=1 x=-\frac{7}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply 2x by 2x+1.

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

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

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

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$4x^{2}+2x-x^{2}+2x-1=6$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+2x+2x-1=6$$

Combine 2x and 2x to get 4x.

$$3x^{2}+4x-1=6$$

Subtract 6 from both sides.

$$3x^{2}+4x-1-6=0$$

Subtract 6 from -1 to get -7.

$$3x^{2}+4x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 4 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-7\right)}}{2\times 3}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-7\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12\left(-7\right)}}{2\times 3}$$

Multiply -12 times -7.

$$x=\frac{-4±\sqrt{16+84}}{2\times 3}$$

Add 16 to 84.

$$x=\frac{-4±\sqrt{100}}{2\times 3}$$

Take the square root of 100.

$$x=\frac{-4±10}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-4±10}{6}$$

Now solve the equation x=\frac{-4±10}{6} when ± is plus. Add -4 to 10.

$$x=\frac{6}{6}$$

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-4±10}{6} when ± is minus. Subtract 10 from -4.

$$x=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{3}$$

The equation is now solved.

$$x=1 x=-\frac{7}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply 2x by 2x+1.

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

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

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

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$4x^{2}+2x-x^{2}+2x-1=6$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+2x+2x-1=6$$

Combine 2x and 2x to get 4x.

$$3x^{2}+4x-1=6$$

Add 1 to both sides.

$$3x^{2}+4x=6+1$$

Add 6 and 1 to get 7.

$$3x^{2}+4x=7$$

Divide both sides by 3.

$$\frac{3x^{2}+4x}{3}=\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{4}{3}x=\frac{7}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{7}{3}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{7}{3}+\frac{4}{9}$$

Add \frac{7}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{25}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{5}{3} x+\frac{2}{3}=-\frac{5}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=1 x=-\frac{7}{3}$$