Solve 12x^2+12=25x

$$12 { x }^{ 2 } +12=25x$$

Solve for x

$$x=\frac{3}{4}=0.75$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Subtract 25x from both sides.

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

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

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

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

$$a+b=-25 ab=12\times 12=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -25.

$$a=-16 b=-9$$

Rewrite 12x^{2}-25x+12 as \left(12x^{2}-16x\right)+\left(-9x+12\right).

$$\left(12x^{2}-16x\right)+\left(-9x+12\right)$$

Factor out 4x in the first and -3 in the second group.

$$4x\left(3x-4\right)-3\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(4x-3\right)$$

To find equation solutions, solve 3x-4=0 and 4x-3=0.

$$x=\frac{4}{3} x=\frac{3}{4}$$

Steps Using the Quadratic Formula

Subtract 25x from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -25 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 12\times 12}}{2\times 12}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 12\times 12}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-25\right)±\sqrt{625-48\times 12}}{2\times 12}$$

Multiply -48 times 12.

$$x=\frac{-\left(-25\right)±\sqrt{625-576}}{2\times 12}$$

Add 625 to -576.

$$x=\frac{-\left(-25\right)±\sqrt{49}}{2\times 12}$$

Take the square root of 49.

$$x=\frac{-\left(-25\right)±7}{2\times 12}$$

The opposite of -25 is 25.

$$x=\frac{25±7}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{25±7}{24}$$

Now solve the equation x=\frac{25±7}{24} when ± is plus. Add 25 to 7.

$$x=\frac{32}{24}$$

Reduce the fraction \frac{32}{24} to lowest terms by extracting and canceling out 8.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{25±7}{24} when ± is minus. Subtract 7 from 25.

$$x=\frac{18}{24}$$

Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.

$$x=\frac{3}{4}$$

The equation is now solved.

$$x=\frac{4}{3} x=\frac{3}{4}$$

Steps for Completing the Square

Subtract 25x from both sides.

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

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 12.

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

Dividing by 12 undoes the multiplication by 12.

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

Divide -12 by 12.

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

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.

$$x^{2}-\frac{25}{12}x+\left(-\frac{25}{24}\right)^{2}=-1+\left(-\frac{25}{24}\right)^{2}$$

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

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

Add -1 to \frac{625}{576}.

$$x^{2}-\frac{25}{12}x+\frac{625}{576}=\frac{49}{576}$$

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}.

$$\left(x-\frac{25}{24}\right)^{2}=\frac{49}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{24}\right)^{2}}=\sqrt{\frac{49}{576}}$$

Simplify.

$$x-\frac{25}{24}=\frac{7}{24} x-\frac{25}{24}=-\frac{7}{24}$$

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

$$x=\frac{4}{3} x=\frac{3}{4}$$