Solve 0=2x^2-20x-42

$$0=2 { x }^{ 2 } -20x-42$$

Solve for x

$$x=\sqrt{46}+5\approx 11.782329983$$

$$x=5-\sqrt{46}\approx -1.782329983$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-20x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -20 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\left(-42\right)}}{2\times 2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 2\left(-42\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-20\right)±\sqrt{400-8\left(-42\right)}}{2\times 2}$$

Multiply -8 times -42.

$$x=\frac{-\left(-20\right)±\sqrt{400+336}}{2\times 2}$$

Add 400 to 336.

$$x=\frac{-\left(-20\right)±\sqrt{736}}{2\times 2}$$

Take the square root of 736.

$$x=\frac{-\left(-20\right)±4\sqrt{46}}{2\times 2}$$

The opposite of -20 is 20.

$$x=\frac{20±4\sqrt{46}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{20±4\sqrt{46}}{4}$$

Now solve the equation x=\frac{20±4\sqrt{46}}{4} when ± is plus. Add 20 to 4\sqrt{46}.

$$x=\frac{4\sqrt{46}+20}{4}$$

Divide 20+4\sqrt{46} by 4.

$$x=\sqrt{46}+5$$

Now solve the equation x=\frac{20±4\sqrt{46}}{4} when ± is minus. Subtract 4\sqrt{46} from 20.

$$x=\frac{20-4\sqrt{46}}{4}$$

Divide 20-4\sqrt{46} by 4.

$$x=5-\sqrt{46}$$

The equation is now solved.

$$x=\sqrt{46}+5 x=5-\sqrt{46}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-20x-42=0$$

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

$$2x^{2}-20x=42$$

Divide both sides by 2.

$$\frac{2x^{2}-20x}{2}=\frac{42}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{20}{2}\right)x=\frac{42}{2}$$

Divide -20 by 2.

$$x^{2}-10x=\frac{42}{2}$$

Divide 42 by 2.

$$x^{2}-10x=21$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=21+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=21+25$$

Add 21 to 25.

$$x^{2}-10x+25=46$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=46$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{46}$$

Simplify.

$$x-5=\sqrt{46} x-5=-\sqrt{46}$$

Add 5 to both sides of the equation.

$$x=\sqrt{46}+5 x=5-\sqrt{46}$$