Solve (2x-1)(2x+1)+(2x-1)(x+2)=x(3+x)-7/4

$$( 2 x - 1 ) ( 2 x + 1 ) + ( 2 x - 1 ) ( x + 2 ) = x ( 3 + x ) - \frac { 7 } { 4 }$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{2}=0.5$$

Steps by Finding Square Root

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(2x\right)^{2}-1+\left(2x-1\right)\left(x+2\right)=x\left(3+x\right)-\frac{7}{4}$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-1+\left(2x-1\right)\left(x+2\right)=x\left(3+x\right)-\frac{7}{4}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-1+\left(2x-1\right)\left(x+2\right)=x\left(3+x\right)-\frac{7}{4}$$

Use the distributive property to multiply 2x-1 by x+2 and combine like terms.

$$4x^{2}-1+2x^{2}+3x-2=x\left(3+x\right)-\frac{7}{4}$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-1+3x-2=x\left(3+x\right)-\frac{7}{4}$$

Subtract 2 from -1 to get -3.

$$6x^{2}-3+3x=x\left(3+x\right)-\frac{7}{4}$$

Use the distributive property to multiply x by 3+x.

$$6x^{2}-3+3x=3x+x^{2}-\frac{7}{4}$$

Subtract 3x from both sides.

$$6x^{2}-3+3x-3x=x^{2}-\frac{7}{4}$$

Combine 3x and -3x to get 0.

$$6x^{2}-3=x^{2}-\frac{7}{4}$$

Subtract x^{2} from both sides.

$$6x^{2}-3-x^{2}=-\frac{7}{4}$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-3=-\frac{7}{4}$$

Add 3 to both sides.

$$5x^{2}=-\frac{7}{4}+3$$

Add -\frac{7}{4} and 3 to get \frac{5}{4}.

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

Divide both sides by 5.

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

Express \frac{\frac{5}{4}}{5} as a single fraction.

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

Cancel out 5 in both numerator and denominator.

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

Take the square root of both sides of the equation.

$$x=\frac{1}{2} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(2x\right)^{2}-1+\left(2x-1\right)\left(x+2\right)=x\left(3+x\right)-\frac{7}{4}$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-1+\left(2x-1\right)\left(x+2\right)=x\left(3+x\right)-\frac{7}{4}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-1+\left(2x-1\right)\left(x+2\right)=x\left(3+x\right)-\frac{7}{4}$$

Use the distributive property to multiply 2x-1 by x+2 and combine like terms.

$$4x^{2}-1+2x^{2}+3x-2=x\left(3+x\right)-\frac{7}{4}$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-1+3x-2=x\left(3+x\right)-\frac{7}{4}$$

Subtract 2 from -1 to get -3.

$$6x^{2}-3+3x=x\left(3+x\right)-\frac{7}{4}$$

Use the distributive property to multiply x by 3+x.

$$6x^{2}-3+3x=3x+x^{2}-\frac{7}{4}$$

Subtract 3x from both sides.

$$6x^{2}-3+3x-3x=x^{2}-\frac{7}{4}$$

Combine 3x and -3x to get 0.

$$6x^{2}-3=x^{2}-\frac{7}{4}$$

Subtract x^{2} from both sides.

$$6x^{2}-3-x^{2}=-\frac{7}{4}$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-3=-\frac{7}{4}$$

Add \frac{7}{4} to both sides.

$$5x^{2}-3+\frac{7}{4}=0$$

Add -3 and \frac{7}{4} to get -\frac{5}{4}.

$$5x^{2}-\frac{5}{4}=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-\frac{5}{4}\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-\frac{5}{4}\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-\frac{5}{4}\right)}}{2\times 5}$$

Multiply -20 times -\frac{5}{4}.

$$x=\frac{0±\sqrt{25}}{2\times 5}$$

Take the square root of 25.

$$x=\frac{0±5}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±5}{10}$$

Now solve the equation x=\frac{0±5}{10} when ± is plus. Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{0±5}{10} when ± is minus. Reduce the fraction \frac{-5}{10} to lowest terms by extracting and canceling out 5.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{1}{2}$$