Solve 2x^2+4x^2-5x+6=0

$$2 x ^ { 2 } + 4 x ^ { 2 } - 5 x + 6 = 0$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{119}i}{12}\approx 0.416666667+0.909059343i$$

$$x=\frac{-\sqrt{119}i+5}{12}\approx 0.416666667-0.909059343i$$

Steps Using the Quadratic Formula

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}-5x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\times 6}}{2\times 6}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-144}}{2\times 6}$$

Add 25 to -144.

$$x=\frac{-\left(-5\right)±\sqrt{-119}}{2\times 6}$$

Take the square root of -119.

$$x=\frac{-\left(-5\right)±\sqrt{119}i}{2\times 6}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{119}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{5±\sqrt{119}i}{12}$$

Now solve the equation x=\frac{5±\sqrt{119}i}{12} when ± is plus. Add 5 to i\sqrt{119}.

$$x=\frac{5+\sqrt{119}i}{12}$$

Now solve the equation x=\frac{5±\sqrt{119}i}{12} when ± is minus. Subtract i\sqrt{119} from 5.

$$x=\frac{-\sqrt{119}i+5}{12}$$

The equation is now solved.

$$x=\frac{5+\sqrt{119}i}{12} x=\frac{-\sqrt{119}i+5}{12}$$

Steps for Completing the Square

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}-5x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 6.

$$\frac{6x^{2}-5x}{6}=-\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{5}{6}x=-\frac{6}{6}$$

Divide -6 by 6.

$$x^{2}-\frac{5}{6}x=-1$$

Divide -\frac{5}{6}, the coefficient of the x term, by 2 to get -\frac{5}{12}. Then add the square of -\frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{6}x+\left(-\frac{5}{12}\right)^{2}=-1+\left(-\frac{5}{12}\right)^{2}$$

Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=-1+\frac{25}{144}$$

Add -1 to \frac{25}{144}.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=-\frac{119}{144}$$

Factor x^{2}-\frac{5}{6}x+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{12}\right)^{2}=-\frac{119}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{12}\right)^{2}}=\sqrt{-\frac{119}{144}}$$

Simplify.

$$x-\frac{5}{12}=\frac{\sqrt{119}i}{12} x-\frac{5}{12}=-\frac{\sqrt{119}i}{12}$$

Add \frac{5}{12} to both sides of the equation.

$$x=\frac{5+\sqrt{119}i}{12} x=\frac{-\sqrt{119}i+5}{12}$$