Solve 20x^2-40-x+10=0

$$20 { x }^{ 2 } -40-x+10=0$$

Solve for x

$$x = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Factoring By Grouping

Add -40 and 10 to get -30.

$$20x^{2}-30-x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$20x^{2}-x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=20\left(-30\right)=-600$$

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 -600.

$$1,-600 2,-300 3,-200 4,-150 5,-120 6,-100 8,-75 10,-60 12,-50 15,-40 20,-30 24,-25$$

Calculate the sum for each pair.

$$1-600=-599 2-300=-298 3-200=-197 4-150=-146 5-120=-115 6-100=-94 8-75=-67 10-60=-50 12-50=-38 15-40=-25 20-30=-10 24-25=-1$$

The solution is the pair that gives sum -1.

$$a=-25 b=24$$

Rewrite 20x^{2}-x-30 as \left(20x^{2}-25x\right)+\left(24x-30\right).

$$\left(20x^{2}-25x\right)+\left(24x-30\right)$$

Factor out 5x in the first and 6 in the second group.

$$5x\left(4x-5\right)+6\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(5x+6\right)$$

To find equation solutions, solve 4x-5=0 and 5x+6=0.

$$x=\frac{5}{4} x=-\frac{6}{5}$$

Steps Using the Quadratic Formula

Add -40 and 10 to get -30.

$$20x^{2}-30-x=0$$

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.

$$20x^{2}-x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 20\left(-30\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-1\right)±\sqrt{1-80\left(-30\right)}}{2\times 20}$$

Multiply -80 times -30.

$$x=\frac{-\left(-1\right)±\sqrt{1+2400}}{2\times 20}$$

Add 1 to 2400.

$$x=\frac{-\left(-1\right)±\sqrt{2401}}{2\times 20}$$

Take the square root of 2401.

$$x=\frac{-\left(-1\right)±49}{2\times 20}$$

The opposite of -1 is 1.

$$x=\frac{1±49}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{1±49}{40}$$

Now solve the equation x=\frac{1±49}{40} when ± is plus. Add 1 to 49.

$$x=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{1±49}{40} when ± is minus. Subtract 49 from 1.

$$x=-\frac{48}{40}$$

Reduce the fraction \frac{-48}{40} to lowest terms by extracting and canceling out 8.

$$x=-\frac{6}{5}$$

The equation is now solved.

$$x=\frac{5}{4} x=-\frac{6}{5}$$

Steps for Completing the Square

Add -40 and 10 to get -30.

$$20x^{2}-30-x=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$20x^{2}-x=30$$

Divide both sides by 20.

$$\frac{20x^{2}-x}{20}=\frac{30}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}-\frac{1}{20}x=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{1}{20}x=\frac{3}{2}$$

Divide -\frac{1}{20}, the coefficient of the x term, by 2 to get -\frac{1}{40}. Then add the square of -\frac{1}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{20}x+\left(-\frac{1}{40}\right)^{2}=\frac{3}{2}+\left(-\frac{1}{40}\right)^{2}$$

Square -\frac{1}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{20}x+\frac{1}{1600}=\frac{3}{2}+\frac{1}{1600}$$

Add \frac{3}{2} to \frac{1}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{20}x+\frac{1}{1600}=\frac{2401}{1600}$$

Factor x^{2}-\frac{1}{20}x+\frac{1}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{40}\right)^{2}=\frac{2401}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{40}\right)^{2}}=\sqrt{\frac{2401}{1600}}$$

Simplify.

$$x-\frac{1}{40}=\frac{49}{40} x-\frac{1}{40}=-\frac{49}{40}$$

Add \frac{1}{40} to both sides of the equation.

$$x=\frac{5}{4} x=-\frac{6}{5}$$