Solve (2x+3)-2x(2x-1)=-x+29

$$(2x+3)-2x(2x-1)=-x+29$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{391}i+5}{8}\approx 0.625-2.471714992i$$

$$x=\frac{5+\sqrt{391}i}{8}\approx 0.625+2.471714992i$$

Steps Using the Quadratic Formula

Add x to both sides.

$$2x+3-2x\left(2x-1\right)+x=29$$

Subtract 29 from both sides.

$$2x+3-2x\left(2x-1\right)+x-29=0$$

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

$$2x+3-4x^{2}+2x+x-29=0$$

Combine 2x and 2x to get 4x.

$$4x+3-4x^{2}+x-29=0$$

Combine 4x and x to get 5x.

$$5x+3-4x^{2}-29=0$$

Subtract 29 from 3 to get -26.

$$5x-26-4x^{2}=0$$

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.

$$-4x^{2}+5x-26=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 5 for b, and -26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-4\right)\left(-26\right)}}{2\left(-4\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-4\right)\left(-26\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-5±\sqrt{25+16\left(-26\right)}}{2\left(-4\right)}$$

Multiply 16 times -26.

$$x=\frac{-5±\sqrt{25-416}}{2\left(-4\right)}$$

Add 25 to -416.

$$x=\frac{-5±\sqrt{-391}}{2\left(-4\right)}$$

Take the square root of -391.

$$x=\frac{-5±\sqrt{391}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-5±\sqrt{391}i}{-8}$$

Now solve the equation x=\frac{-5±\sqrt{391}i}{-8} when ± is plus. Add -5 to i\sqrt{391}.

$$x=\frac{-5+\sqrt{391}i}{-8}$$

Divide -5+i\sqrt{391} by -8.

$$x=\frac{-\sqrt{391}i+5}{8}$$

Now solve the equation x=\frac{-5±\sqrt{391}i}{-8} when ± is minus. Subtract i\sqrt{391} from -5.

$$x=\frac{-\sqrt{391}i-5}{-8}$$

Divide -5-i\sqrt{391} by -8.

$$x=\frac{5+\sqrt{391}i}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{391}i+5}{8} x=\frac{5+\sqrt{391}i}{8}$$

Steps for Completing the Square

Add x to both sides.

$$2x+3-2x\left(2x-1\right)+x=29$$

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

$$2x+3-4x^{2}+2x+x=29$$

Combine 2x and 2x to get 4x.

$$4x+3-4x^{2}+x=29$$

Combine 4x and x to get 5x.

$$5x+3-4x^{2}=29$$

Subtract 3 from both sides.

$$5x-4x^{2}=29-3$$

Subtract 3 from 29 to get 26.

$$5x-4x^{2}=26$$

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.

$$-4x^{2}+5x=26$$

Divide both sides by -4.

$$\frac{-4x^{2}+5x}{-4}=\frac{26}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{5}{-4}x=\frac{26}{-4}$$

Divide 5 by -4.

$$x^{2}-\frac{5}{4}x=\frac{26}{-4}$$

Reduce the fraction \frac{26}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{4}x=-\frac{13}{2}$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=-\frac{13}{2}+\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{13}{2}+\frac{25}{64}$$

Add -\frac{13}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{391}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{8}\right)^{2}=-\frac{391}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{-\frac{391}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{\sqrt{391}i}{8} x-\frac{5}{8}=-\frac{\sqrt{391}i}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$x=\frac{5+\sqrt{391}i}{8} x=\frac{-\sqrt{391}i+5}{8}$$