Solve 18x^2-35x+12=0

$$18 { x }^{ 2 } -35x+12=0$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=\frac{4}{9}\approx 0.444444444$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 18x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-35 ab=18\times 12=216$$

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 216.

$$-1,-216 -2,-108 -3,-72 -4,-54 -6,-36 -8,-27 -9,-24 -12,-18$$

Calculate the sum for each pair.

$$-1-216=-217 -2-108=-110 -3-72=-75 -4-54=-58 -6-36=-42 -8-27=-35 -9-24=-33 -12-18=-30$$

The solution is the pair that gives sum -35.

$$a=-27 b=-8$$

Rewrite 18x^{2}-35x+12 as \left(18x^{2}-27x\right)+\left(-8x+12\right).

$$\left(18x^{2}-27x\right)+\left(-8x+12\right)$$

Factor out 9x in the first and -4 in the second group.

$$9x\left(2x-3\right)-4\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(9x-4\right)$$

To find equation solutions, solve 2x-3=0 and 9x-4=0.

$$x=\frac{3}{2} x=\frac{4}{9}$$

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.

$$18x^{2}-35x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -35 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 18\times 12}}{2\times 18}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 18\times 12}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-35\right)±\sqrt{1225-72\times 12}}{2\times 18}$$

Multiply -72 times 12.

$$x=\frac{-\left(-35\right)±\sqrt{1225-864}}{2\times 18}$$

Add 1225 to -864.

$$x=\frac{-\left(-35\right)±\sqrt{361}}{2\times 18}$$

Take the square root of 361.

$$x=\frac{-\left(-35\right)±19}{2\times 18}$$

The opposite of -35 is 35.

$$x=\frac{35±19}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{35±19}{36}$$

Now solve the equation x=\frac{35±19}{36} when ± is plus. Add 35 to 19.

$$x=\frac{54}{36}$$

Reduce the fraction \frac{54}{36} to lowest terms by extracting and canceling out 18.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{35±19}{36} when ± is minus. Subtract 19 from 35.

$$x=\frac{16}{36}$$

Reduce the fraction \frac{16}{36} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{9}$$

The equation is now solved.

$$x=\frac{3}{2} x=\frac{4}{9}$$

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.

$$18x^{2}-35x+12=0$$

Subtract 12 from both sides of the equation.

$$18x^{2}-35x+12-12=-12$$

Subtracting 12 from itself leaves 0.

$$18x^{2}-35x=-12$$

Divide both sides by 18.

$$\frac{18x^{2}-35x}{18}=-\frac{12}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}-\frac{35}{18}x=-\frac{12}{18}$$

Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{35}{18}x=-\frac{2}{3}$$

Divide -\frac{35}{18}, the coefficient of the x term, by 2 to get -\frac{35}{36}. Then add the square of -\frac{35}{36} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{18}x+\left(-\frac{35}{36}\right)^{2}=-\frac{2}{3}+\left(-\frac{35}{36}\right)^{2}$$

Square -\frac{35}{36} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{18}x+\frac{1225}{1296}=-\frac{2}{3}+\frac{1225}{1296}$$

Add -\frac{2}{3} to \frac{1225}{1296} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{35}{18}x+\frac{1225}{1296}=\frac{361}{1296}$$

Factor x^{2}-\frac{35}{18}x+\frac{1225}{1296}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{35}{36}\right)^{2}=\frac{361}{1296}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{36}\right)^{2}}=\sqrt{\frac{361}{1296}}$$

Simplify.

$$x-\frac{35}{36}=\frac{19}{36} x-\frac{35}{36}=-\frac{19}{36}$$

Add \frac{35}{36} to both sides of the equation.

$$x=\frac{3}{2} x=\frac{4}{9}$$