Solve 13x=12x^2+3

$$13 x = 12 x ^ { 2 } + 3$$

Solve for x

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

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Subtract 12x^{2} from both sides.

$$13x-12x^{2}=3$$

Subtract 3 from both sides.

$$13x-12x^{2}-3=0$$

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

$$-12x^{2}+13x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -12x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=13 ab=-12\left(-3\right)=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=9 b=4$$

Rewrite -12x^{2}+13x-3 as \left(-12x^{2}+9x\right)+\left(4x-3\right).

$$\left(-12x^{2}+9x\right)+\left(4x-3\right)$$

Factor out 3x in the first and -1 in the second group.

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

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

$$\left(-4x+3\right)\left(3x-1\right)$$

To find equation solutions, solve -4x+3=0 and 3x-1=0.

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

Steps Using the Quadratic Formula

Subtract 12x^{2} from both sides.

$$13x-12x^{2}=3$$

Subtract 3 from both sides.

$$13x-12x^{2}-3=0$$

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.

$$-12x^{2}+13x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 13 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-12\right)\left(-3\right)}}{2\left(-12\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-12\right)\left(-3\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-13±\sqrt{169+48\left(-3\right)}}{2\left(-12\right)}$$

Multiply 48 times -3.

$$x=\frac{-13±\sqrt{169-144}}{2\left(-12\right)}$$

Add 169 to -144.

$$x=\frac{-13±\sqrt{25}}{2\left(-12\right)}$$

Take the square root of 25.

$$x=\frac{-13±5}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-13±5}{-24}$$

Now solve the equation x=\frac{-13±5}{-24} when ± is plus. Add -13 to 5.

$$x=-\frac{8}{-24}$$

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-13±5}{-24} when ± is minus. Subtract 5 from -13.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 12x^{2} from both sides.

$$13x-12x^{2}=3$$

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.

$$-12x^{2}+13x=3$$

Divide both sides by -12.

$$\frac{-12x^{2}+13x}{-12}=\frac{3}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{13}{-12}x=\frac{3}{-12}$$

Divide 13 by -12.

$$x^{2}-\frac{13}{12}x=\frac{3}{-12}$$

Reduce the fraction \frac{3}{-12} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{13}{12}x=-\frac{1}{4}$$

Divide -\frac{13}{12}, the coefficient of the x term, by 2 to get -\frac{13}{24}. Then add the square of -\frac{13}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{12}x+\left(-\frac{13}{24}\right)^{2}=-\frac{1}{4}+\left(-\frac{13}{24}\right)^{2}$$

Square -\frac{13}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{12}x+\frac{169}{576}=-\frac{1}{4}+\frac{169}{576}$$

Add -\frac{1}{4} to \frac{169}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{12}x+\frac{169}{576}=\frac{25}{576}$$

Factor x^{2}-\frac{13}{12}x+\frac{169}{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}.

$$\left(x-\frac{13}{24}\right)^{2}=\frac{25}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{24}\right)^{2}}=\sqrt{\frac{25}{576}}$$

Simplify.

$$x-\frac{13}{24}=\frac{5}{24} x-\frac{13}{24}=-\frac{5}{24}$$

Add \frac{13}{24} to both sides of the equation.

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