Solve 24x^2-50x+26=0

$$24 { x }^{ 2 } -50x+26=0$$

Solve for x

$$x=1$$

$$x = \frac{13}{12} = 1\frac{1}{12} \approx 1.083333333$$

Steps Using Factoring By Grouping

Divide both sides by 2.

$$12x^{2}-25x+13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+13. To find a and b, set up a system to be solved.

$$a+b=-25 ab=12\times 13=156$$

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 156.

$$-1,-156 -2,-78 -3,-52 -4,-39 -6,-26 -12,-13$$

Calculate the sum for each pair.

$$-1-156=-157 -2-78=-80 -3-52=-55 -4-39=-43 -6-26=-32 -12-13=-25$$

The solution is the pair that gives sum -25.

$$a=-13 b=-12$$

Rewrite 12x^{2}-25x+13 as \left(12x^{2}-13x\right)+\left(-12x+13\right).

$$\left(12x^{2}-13x\right)+\left(-12x+13\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(12x-13\right)-\left(12x-13\right)$$

Factor out common term 12x-13 by using distributive property.

$$\left(12x-13\right)\left(x-1\right)$$

To find equation solutions, solve 12x-13=0 and x-1=0.

$$x=\frac{13}{12} x=1$$

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.

$$24x^{2}-50x+26=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, -50 for b, and 26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 24\times 26}}{2\times 24}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 24\times 26}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-50\right)±\sqrt{2500-96\times 26}}{2\times 24}$$

Multiply -96 times 26.

$$x=\frac{-\left(-50\right)±\sqrt{2500-2496}}{2\times 24}$$

Add 2500 to -2496.

$$x=\frac{-\left(-50\right)±\sqrt{4}}{2\times 24}$$

Take the square root of 4.

$$x=\frac{-\left(-50\right)±2}{2\times 24}$$

The opposite of -50 is 50.

$$x=\frac{50±2}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{50±2}{48}$$

Now solve the equation x=\frac{50±2}{48} when ± is plus. Add 50 to 2.

$$x=\frac{52}{48}$$

Reduce the fraction \frac{52}{48} to lowest terms by extracting and canceling out 4.

$$x=\frac{13}{12}$$

Now solve the equation x=\frac{50±2}{48} when ± is minus. Subtract 2 from 50.

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

Divide 48 by 48.

$$x=1$$

The equation is now solved.

$$x=\frac{13}{12} x=1$$

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.

$$24x^{2}-50x+26=0$$

Subtract 26 from both sides of the equation.

$$24x^{2}-50x+26-26=-26$$

Subtracting 26 from itself leaves 0.

$$24x^{2}-50x=-26$$

Divide both sides by 24.

$$\frac{24x^{2}-50x}{24}=-\frac{26}{24}$$

Dividing by 24 undoes the multiplication by 24.

$$x^{2}+\left(-\frac{50}{24}\right)x=-\frac{26}{24}$$

Reduce the fraction \frac{-50}{24} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{25}{12}x=-\frac{26}{24}$$

Reduce the fraction \frac{-26}{24} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{25}{12}x=-\frac{13}{12}$$

Divide -\frac{25}{12}, the coefficient of the x term, by 2 to get -\frac{25}{24}. Then add the square of -\frac{25}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{12}x+\left(-\frac{25}{24}\right)^{2}=-\frac{13}{12}+\left(-\frac{25}{24}\right)^{2}$$

Square -\frac{25}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{12}x+\frac{625}{576}=-\frac{13}{12}+\frac{625}{576}$$

Add -\frac{13}{12} to \frac{625}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{25}{12}x+\frac{625}{576}=\frac{1}{576}$$

Factor x^{2}-\frac{25}{12}x+\frac{625}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{24}\right)^{2}=\frac{1}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{24}\right)^{2}}=\sqrt{\frac{1}{576}}$$

Simplify.

$$x-\frac{25}{24}=\frac{1}{24} x-\frac{25}{24}=-\frac{1}{24}$$

Add \frac{25}{24} to both sides of the equation.

$$x=\frac{13}{12} x=1$$