Solve 10x^2+20x-71=9

$$10 x ^ { 2 } + 20 x - 71 = 9$$

Solve for x

$$x=-4$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 9 from both sides.

$$10x^{2}+20x-71-9=0$$

Subtract 9 from -71 to get -80.

$$10x^{2}+20x-80=0$$

Divide both sides by 10.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite x^{2}+2x-8 as \left(x^{2}-2x\right)+\left(4x-8\right).

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

Factor out x in the first and 4 in the second group.

$$x\left(x-2\right)+4\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+4\right)$$

To find equation solutions, solve x-2=0 and x+4=0.

$$x=2 x=-4$$

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.

$$10x^{2}+20x-71=9$$

Subtract 9 from both sides of the equation.

$$10x^{2}+20x-71-9=9-9$$

Subtracting 9 from itself leaves 0.

$$10x^{2}+20x-71-9=0$$

Subtract 9 from -71.

$$10x^{2}+20x-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 20 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 10\left(-80\right)}}{2\times 10}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 10\left(-80\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-20±\sqrt{400-40\left(-80\right)}}{2\times 10}$$

Multiply -40 times -80.

$$x=\frac{-20±\sqrt{400+3200}}{2\times 10}$$

Add 400 to 3200.

$$x=\frac{-20±\sqrt{3600}}{2\times 10}$$

Take the square root of 3600.

$$x=\frac{-20±60}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-20±60}{20}$$

Now solve the equation x=\frac{-20±60}{20} when ± is plus. Add -20 to 60.

$$x=\frac{40}{20}$$

Divide 40 by 20.

$$x=2$$

Now solve the equation x=\frac{-20±60}{20} when ± is minus. Subtract 60 from -20.

$$x=-\frac{80}{20}$$

Divide -80 by 20.

$$x=-4$$

The equation is now solved.

$$x=2 x=-4$$

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.

$$10x^{2}+20x-71=9$$

Add 71 to both sides of the equation.

$$10x^{2}+20x-71-\left(-71\right)=9-\left(-71\right)$$

Subtracting -71 from itself leaves 0.

$$10x^{2}+20x=9-\left(-71\right)$$

Subtract -71 from 9.

$$10x^{2}+20x=80$$

Divide both sides by 10.

$$\frac{10x^{2}+20x}{10}=\frac{80}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{20}{10}x=\frac{80}{10}$$

Divide 20 by 10.

$$x^{2}+2x=\frac{80}{10}$$

Divide 80 by 10.

$$x^{2}+2x=8$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=8+1^{2}$$

Square 1.

$$x^{2}+2x+1=8+1$$

Add 8 to 1.

$$x^{2}+2x+1=9$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+1=3 x+1=-3$$

Subtract 1 from both sides of the equation.

$$x=2 x=-4$$