Solve x^2+2x+3=4x^2+5x+6

$${ x }^{ 2 } +2x+3=4 { x }^{ 2 } +5x+6$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{3}i}{2}\approx -0.5+0.866025404i$$

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$x^{2}+2x+3-4x^{2}=5x+6$$

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

$$-3x^{2}+2x+3=5x+6$$

Subtract 5x from both sides.

$$-3x^{2}+2x+3-5x=6$$

Combine 2x and -5x to get -3x.

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

Subtract 6 from both sides.

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

Subtract 6 from 3 to get -3.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9+12\left(-3\right)}}{2\left(-3\right)}$$

Multiply 12 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-36}}{2\left(-3\right)}$$

Add 9 to -36.

$$x=\frac{-\left(-3\right)±\sqrt{-27}}{2\left(-3\right)}$$

Take the square root of -27.

$$x=\frac{-\left(-3\right)±3\sqrt{3}i}{2\left(-3\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{3}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{3±3\sqrt{3}i}{-6}$$

Now solve the equation x=\frac{3±3\sqrt{3}i}{-6} when ± is plus. Add 3 to 3i\sqrt{3}.

$$x=\frac{3+3\sqrt{3}i}{-6}$$

Divide 3+3i\sqrt{3} by -6.

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

Now solve the equation x=\frac{3±3\sqrt{3}i}{-6} when ± is minus. Subtract 3i\sqrt{3} from 3.

$$x=\frac{-3\sqrt{3}i+3}{-6}$$

Divide 3-3i\sqrt{3} by -6.

$$x=\frac{-1+\sqrt{3}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{3}i-1}{2} x=\frac{-1+\sqrt{3}i}{2}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$x^{2}+2x+3-4x^{2}=5x+6$$

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

$$-3x^{2}+2x+3=5x+6$$

Subtract 5x from both sides.

$$-3x^{2}+2x+3-5x=6$$

Combine 2x and -5x to get -3x.

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

Subtract 3 from both sides.

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

Subtract 3 from 6 to get 3.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -3 by -3.

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

Divide 3 by -3.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -1 to \frac{1}{4}.

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

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{3}i}{2} x+\frac{1}{2}=-\frac{\sqrt{3}i}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{-1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-1}{2}$$