Solve 3x^2+5=4x

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

Solve for x (complex solution)

$$x=\frac{2+\sqrt{11}i}{3}\approx 0.666666667+1.105541597i$$

$$x=\frac{-\sqrt{11}i+2}{3}\approx 0.666666667-1.105541597i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

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

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.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-60}}{2\times 3}$$

Add 16 to -60.

$$x=\frac{-\left(-4\right)±\sqrt{-44}}{2\times 3}$$

Take the square root of -44.

$$x=\frac{-\left(-4\right)±2\sqrt{11}i}{2\times 3}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{11}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±2\sqrt{11}i}{6}$$

Now solve the equation x=\frac{4±2\sqrt{11}i}{6} when ± is plus. Add 4 to 2i\sqrt{11}.

$$x=\frac{4+2\sqrt{11}i}{6}$$

Divide 4+2i\sqrt{11} by 6.

$$x=\frac{2+\sqrt{11}i}{3}$$

Now solve the equation x=\frac{4±2\sqrt{11}i}{6} when ± is minus. Subtract 2i\sqrt{11} from 4.

$$x=\frac{-2\sqrt{11}i+4}{6}$$

Divide 4-2i\sqrt{11} by 6.

$$x=\frac{-\sqrt{11}i+2}{3}$$

The equation is now solved.

$$x=\frac{2+\sqrt{11}i}{3} x=\frac{-\sqrt{11}i+2}{3}$$

Steps for Completing the Square

Subtract 4x from both sides.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{5}{3}+\frac{4}{9}$$

Add -\frac{5}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{11}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=-\frac{11}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{-\frac{11}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{11}i}{3} x-\frac{2}{3}=-\frac{\sqrt{11}i}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{2+\sqrt{11}i}{3} x=\frac{-\sqrt{11}i+2}{3}$$