Solve x^2-8x-1280=0

$${ x }^{ 2 } -8x-1280=0$$

Solve for x

$$x=-32$$

$$x=40$$

Steps Using Factoring

To solve the equation, factor x^{2}-8x-1280 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=-1280$$

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 -1280.

$$1,-1280 2,-640 4,-320 5,-256 8,-160 10,-128 16,-80 20,-64 32,-40$$

Calculate the sum for each pair.

$$1-1280=-1279 2-640=-638 4-320=-316 5-256=-251 8-160=-152 10-128=-118 16-80=-64 20-64=-44 32-40=-8$$

The solution is the pair that gives sum -8.

$$a=-40 b=32$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-40\right)\left(x+32\right)$$

To find equation solutions, solve x-40=0 and x+32=0.

$$x=40 x=-32$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-1280. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-1280\right)=-1280$$

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 -1280.

$$1,-1280 2,-640 4,-320 5,-256 8,-160 10,-128 16,-80 20,-64 32,-40$$

Calculate the sum for each pair.

$$1-1280=-1279 2-640=-638 4-320=-316 5-256=-251 8-160=-152 10-128=-118 16-80=-64 20-64=-44 32-40=-8$$

The solution is the pair that gives sum -8.

$$a=-40 b=32$$

Rewrite x^{2}-8x-1280 as \left(x^{2}-40x\right)+\left(32x-1280\right).

$$\left(x^{2}-40x\right)+\left(32x-1280\right)$$

Factor out x in the first and 32 in the second group.

$$x\left(x-40\right)+32\left(x-40\right)$$

Factor out common term x-40 by using distributive property.

$$\left(x-40\right)\left(x+32\right)$$

To find equation solutions, solve x-40=0 and x+32=0.

$$x=40 x=-32$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}-8x-1280=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -1280 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1280\right)}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-1280\right)}}{2}$$

Multiply -4 times -1280.

$$x=\frac{-\left(-8\right)±\sqrt{64+5120}}{2}$$

Add 64 to 5120.

$$x=\frac{-\left(-8\right)±\sqrt{5184}}{2}$$

Take the square root of 5184.

$$x=\frac{-\left(-8\right)±72}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±72}{2}$$

Now solve the equation x=\frac{8±72}{2} when ± is plus. Add 8 to 72.

$$x=\frac{80}{2}$$

Divide 80 by 2.

$$x=40$$

Now solve the equation x=\frac{8±72}{2} when ± is minus. Subtract 72 from 8.

$$x=-\frac{64}{2}$$

Divide -64 by 2.

$$x=-32$$

The equation is now solved.

$$x=40 x=-32$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}-8x-1280=0$$

Add 1280 to both sides of the equation.

$$x^{2}-8x-1280-\left(-1280\right)=-\left(-1280\right)$$

Subtracting -1280 from itself leaves 0.

$$x^{2}-8x=-\left(-1280\right)$$

Subtract -1280 from 0.

$$x^{2}-8x=1280$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=1280+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=1280+16$$

Add 1280 to 16.

$$x^{2}-8x+16=1296$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=1296$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{1296}$$

Simplify.

$$x-4=36 x-4=-36$$

Add 4 to both sides of the equation.

$$x=40 x=-32$$