Solve 2x+1=4x^2+4x+5

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{15}i}{4}\approx -0.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i-1}{4}\approx -0.25-0.968245837i$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

Subtract 5 from both sides.

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

Subtract 5 from 1 to get -4.

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

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.

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

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

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

Square -2.

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

Multiply -4 times -4.

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

Multiply 16 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4-64}}{2\left(-4\right)}$$

Add 4 to -64.

$$x=\frac{-\left(-2\right)±\sqrt{-60}}{2\left(-4\right)}$$

Take the square root of -60.

$$x=\frac{-\left(-2\right)±2\sqrt{15}i}{2\left(-4\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{15}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{2±2\sqrt{15}i}{-8}$$

Now solve the equation x=\frac{2±2\sqrt{15}i}{-8} when ± is plus. Add 2 to 2i\sqrt{15}.

$$x=\frac{2+2\sqrt{15}i}{-8}$$

Divide 2+2i\sqrt{15} by -8.

$$x=\frac{-\sqrt{15}i-1}{4}$$

Now solve the equation x=\frac{2±2\sqrt{15}i}{-8} when ± is minus. Subtract 2i\sqrt{15} from 2.

$$x=\frac{-2\sqrt{15}i+2}{-8}$$

Divide 2-2i\sqrt{15} by -8.

$$x=\frac{-1+\sqrt{15}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{15}i-1}{4} x=\frac{-1+\sqrt{15}i}{4}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

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

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

Subtract 1 from both sides.

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

Subtract 1 from 5 to get 4.

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

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.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

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

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

Divide 4 by -4.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-1+\frac{1}{16}$$

Add -1 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{15}i}{4} x+\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{-1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i-1}{4}$$