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

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

Solve for x

$$x=-7$$

$$x=-3$$

Steps Using Factoring By Grouping

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

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

Divide both sides by 2.

$$-x^{2}-10x-21=0$$

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.

$$a+b=-10 ab=-\left(-21\right)=21$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.

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

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-3 b=-7$$

Rewrite -x^{2}-10x-21 as \left(-x^{2}-3x\right)+\left(-7x-21\right).

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

Factor out x in the first and 7 in the second group.

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

Factor out common term -x-3 by using distributive property.

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

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

$$x=-3 x=-7$$

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}.

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

Square -20.

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

Multiply -4 times -2.

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

Multiply 8 times -42.

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

Add 400 to -336.

$$x=\frac{-\left(-20\right)±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

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

The opposite of -20 is 20.

$$x=\frac{20±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{20±8}{-4}$$

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

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

Divide 28 by -4.

$$x=-7$$

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

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

Divide 12 by -4.

$$x=-3$$

The equation is now solved.

$$x=-7 x=-3$$

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.

$$x^{2}+10x+5^{2}=-21+5^{2}$$

Square 5.

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

Add -21 to 25.

$$x^{2}+10x+25=4$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{4}$$

Simplify.

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

Subtract 5 from both sides of the equation.

$$x=-3 x=-7$$