Solve 5x^2-25x=5x-40

$$5 x ^ { 2 } - 25 x = 5 x - 40$$

Solve for x

$$x=2$$

$$x=4$$

Steps Using Factoring By Grouping

Subtract 5x from both sides.

$$5x^{2}-25x-5x=-40$$

Combine -25x and -5x to get -30x.

$$5x^{2}-30x=-40$$

Add 40 to both sides.

$$5x^{2}-30x+40=0$$

Divide both sides by 5.

$$x^{2}-6x+8=0$$

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.

$$a+b=-6 ab=1\times 8=8$$

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 8.

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

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite x^{2}-6x+8 as \left(x^{2}-4x\right)+\left(-2x+8\right).

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

Factor out x in the first and -2 in the second group.

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

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

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

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

$$x=4 x=2$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$5x^{2}-25x-5x=-40$$

Combine -25x and -5x to get -30x.

$$5x^{2}-30x=-40$$

Add 40 to both sides.

$$5x^{2}-30x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -30 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 5\times 40}}{2\times 5}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 5\times 40}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-30\right)±\sqrt{900-20\times 40}}{2\times 5}$$

Multiply -20 times 40.

$$x=\frac{-\left(-30\right)±\sqrt{900-800}}{2\times 5}$$

Add 900 to -800.

$$x=\frac{-\left(-30\right)±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$x=\frac{-\left(-30\right)±10}{2\times 5}$$

The opposite of -30 is 30.

$$x=\frac{30±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{30±10}{10}$$

Now solve the equation x=\frac{30±10}{10} when ± is plus. Add 30 to 10.

$$x=\frac{40}{10}$$

Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{30±10}{10} when ± is minus. Subtract 10 from 30.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

The equation is now solved.

$$x=4 x=2$$

Steps for Completing the Square

Subtract 5x from both sides.

$$5x^{2}-25x-5x=-40$$

Combine -25x and -5x to get -30x.

$$5x^{2}-30x=-40$$

Divide both sides by 5.

$$\frac{5x^{2}-30x}{5}=-\frac{40}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{30}{5}\right)x=-\frac{40}{5}$$

Divide -30 by 5.

$$x^{2}-6x=-\frac{40}{5}$$

Divide -40 by 5.

$$x^{2}-6x=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-8+9$$

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{1}$$

Simplify.

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

Add 3 to both sides of the equation.

$$x=4 x=2$$