Solve 25x^2-8x=12x-4

$$25 { x }^{ 2 } -8x=12x-4$$

Solve for x

$$x=\frac{2}{5}=0.4$$

Steps Using Factoring By Grouping

Subtract 12x from both sides.

$$25x^{2}-8x-12x=-4$$

Combine -8x and -12x to get -20x.

$$25x^{2}-20x=-4$$

Add 4 to both sides.

$$25x^{2}-20x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-20 ab=25\times 4=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

$$a=-10 b=-10$$

Rewrite 25x^{2}-20x+4 as \left(25x^{2}-10x\right)+\left(-10x+4\right).

$$\left(25x^{2}-10x\right)+\left(-10x+4\right)$$

Factor out 5x in the first and -2 in the second group.

$$5x\left(5x-2\right)-2\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(5x-2\right)$$

Rewrite as a binomial square.

$$\left(5x-2\right)^{2}$$

To find equation solution, solve 5x-2=0.

$$x=\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 12x from both sides.

$$25x^{2}-8x-12x=-4$$

Combine -8x and -12x to get -20x.

$$25x^{2}-20x=-4$$

Add 4 to both sides.

$$25x^{2}-20x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -20 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 25\times 4}}{2\times 25}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 25\times 4}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-20\right)±\sqrt{400-100\times 4}}{2\times 25}$$

Multiply -100 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-400}}{2\times 25}$$

Add 400 to -400.

$$x=\frac{-\left(-20\right)±\sqrt{0}}{2\times 25}$$

Take the square root of 0.

$$x=-\frac{-20}{2\times 25}$$

The opposite of -20 is 20.

$$x=\frac{20}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{20}{50}$$

Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{2}{5}$$

Steps for Completing the Square

Subtract 12x from both sides.

$$25x^{2}-8x-12x=-4$$

Combine -8x and -12x to get -20x.

$$25x^{2}-20x=-4$$

Divide both sides by 25.

$$\frac{25x^{2}-20x}{25}=-\frac{4}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{20}{25}\right)x=-\frac{4}{25}$$

Reduce the fraction \frac{-20}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{4}{5}x=-\frac{4}{25}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-\frac{4}{25}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{-4+4}{25}$$

Add -\frac{4}{25} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=0$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{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}.

$$\left(x-\frac{2}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{2}{5}=0 x-\frac{2}{5}=0$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{2}{5} x=\frac{2}{5}$$

The equation is now solved. Solutions are the same.

$$x=\frac{2}{5}$$