Solve (2x+1)(1-x-3)+4x^2-1=0

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

Solve for x

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

$$x=3$$

Steps Using Factoring By Grouping

Subtract 3 from 1 to get -2.

$$\left(2x+1\right)\left(-2-x\right)+4x^{2}-1=0$$

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

$$-5x-2x^{2}-2+4x^{2}-1=0$$

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

$$-5x+2x^{2}-2-1=0$$

Subtract 1 from -2 to get -3.

$$-5x+2x^{2}-3=0$$

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

$$2x^{2}-5x-3=0$$

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

$$a+b=-5 ab=2\left(-3\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 2x^{2}-5x-3 as \left(2x^{2}-6x\right)+\left(x-3\right).

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

Factor out 2x in 2x^{2}-6x.

$$2x\left(x-3\right)+x-3$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x+1\right)$$

To find equation solutions, solve x-3=0 and 2x+1=0.

$$x=3 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 3 from 1 to get -2.

$$\left(2x+1\right)\left(-2-x\right)+4x^{2}-1=0$$

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

$$-5x-2x^{2}-2+4x^{2}-1=0$$

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

$$-5x+2x^{2}-2-1=0$$

Subtract 1 from -2 to get -3.

$$-5x+2x^{2}-3=0$$

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.

$$2x^{2}-5x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 2}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±7}{4}$$

Now solve the equation x=\frac{5±7}{4} when ± is plus. Add 5 to 7.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{5±7}{4} when ± is minus. Subtract 7 from 5.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=3 x=-\frac{1}{2}$$

Steps for Completing the Square

Subtract 3 from 1 to get -2.

$$\left(2x+1\right)\left(-2-x\right)+4x^{2}-1=0$$

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

$$-5x-2x^{2}-2+4x^{2}-1=0$$

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

$$-5x+2x^{2}-2-1=0$$

Subtract 1 from -2 to get -3.

$$-5x+2x^{2}-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-5x+2x^{2}=3$$

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.

$$2x^{2}-5x=3$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{7}{4} x-\frac{5}{4}=-\frac{7}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=3 x=-\frac{1}{2}$$