Solve x^2-4x+2x^2-42=0

$$x ^ { 2 } - 4 x + 2 x ^ { 2 } - 42 = 0$$

Solve for x

$$x = \frac{\sqrt{130} + 2}{3} \approx 4.467251417$$

$$x=\frac{2-\sqrt{130}}{3}\approx -3.133918084$$

Steps Using the Quadratic Formula

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

$$3x^{2}-4x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -4 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\left(-42\right)}}{2\times 3}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3\left(-42\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12\left(-42\right)}}{2\times 3}$$

Multiply -12 times -42.

$$x=\frac{-\left(-4\right)±\sqrt{16+504}}{2\times 3}$$

Add 16 to 504.

$$x=\frac{-\left(-4\right)±\sqrt{520}}{2\times 3}$$

Take the square root of 520.

$$x=\frac{-\left(-4\right)±2\sqrt{130}}{2\times 3}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{130}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±2\sqrt{130}}{6}$$

Now solve the equation x=\frac{4±2\sqrt{130}}{6} when ± is plus. Add 4 to 2\sqrt{130}.

$$x=\frac{2\sqrt{130}+4}{6}$$

Divide 4+2\sqrt{130} by 6.

$$x=\frac{\sqrt{130}+2}{3}$$

Now solve the equation x=\frac{4±2\sqrt{130}}{6} when ± is minus. Subtract 2\sqrt{130} from 4.

$$x=\frac{4-2\sqrt{130}}{6}$$

Divide 4-2\sqrt{130} by 6.

$$x=\frac{2-\sqrt{130}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{130}+2}{3} x=\frac{2-\sqrt{130}}{3}$$

Steps for Completing the Square

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

$$3x^{2}-4x-42=0$$

Add 42 to both sides. Anything plus zero gives itself.

$$3x^{2}-4x=42$$

Divide both sides by 3.

$$\frac{3x^{2}-4x}{3}=\frac{42}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{4}{3}x=\frac{42}{3}$$

Divide 42 by 3.

$$x^{2}-\frac{4}{3}x=14$$

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.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=14+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=14+\frac{4}{9}$$

Add 14 to \frac{4}{9}.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{130}{9}$$

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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{130}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{130}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{130}}{3} x-\frac{2}{3}=-\frac{\sqrt{130}}{3}$$

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

$$x=\frac{\sqrt{130}+2}{3} x=\frac{2-\sqrt{130}}{3}$$