Solve 2x^2-8x-2*21=0

$$2 { x }^{ 2 } -8x-2 \times 21 = 0$$

Solve for x

$$x=-3$$

$$x=7$$

Steps Using Factoring By Grouping

Multiply 2 and 21 to get 42.

$$2x^{2}-8x-42=0$$

Divide both sides by 2.

$$x^{2}-4x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-21\right)=-21$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -21.

$$1,-21 3,-7$$

Calculate the sum for each pair.

$$1-21=-20 3-7=-4$$

The solution is the pair that gives sum -4.

$$a=-7 b=3$$

Rewrite x^{2}-4x-21 as \left(x^{2}-7x\right)+\left(3x-21\right).

$$\left(x^{2}-7x\right)+\left(3x-21\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-7\right)+3\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+3\right)$$

To find equation solutions, solve x-7=0 and x+3=0.

$$x=7 x=-3$$

Steps Using the Quadratic Formula

Multiply 2 and 21 to get 42.

$$2x^{2}-8x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-42\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-42\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-42\right)}}{2\times 2}$$

Multiply -8 times -42.

$$x=\frac{-\left(-8\right)±\sqrt{64+336}}{2\times 2}$$

Add 64 to 336.

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

Take the square root of 400.

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

The opposite of -8 is 8.

$$x=\frac{8±20}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±20}{4}$$

Now solve the equation x=\frac{8±20}{4} when ± is plus. Add 8 to 20.

$$x=\frac{28}{4}$$

Divide 28 by 4.

$$x=7$$

Now solve the equation x=\frac{8±20}{4} when ± is minus. Subtract 20 from 8.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=7 x=-3$$

Steps for Completing the Square

Multiply 2 and 21 to get 42.

$$2x^{2}-8x-42=0$$

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

$$2x^{2}-8x=42$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{42}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{42}{2}$$

Divide -8 by 2.

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

Divide 42 by 2.

$$x^{2}-4x=21$$

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

$$x^{2}-4x+\left(-2\right)^{2}=21+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=21+4$$

Add 21 to 4.

$$x^{2}-4x+4=25$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-2=5 x-2=-5$$

Add 2 to both sides of the equation.

$$x=7 x=-3$$