Solve 6x^2=17x-12

$$6 x ^ { 2 } = 17 x - 12$$

Solve for x

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

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

Steps Using Factoring By Grouping

Subtract 17x from both sides.

$$6x^{2}-17x=-12$$

Add 12 to both sides.

$$6x^{2}-17x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-17 ab=6\times 12=72$$

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 72.

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

Calculate the sum for each pair.

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

The solution is the pair that gives sum -17.

$$a=-9 b=-8$$

Rewrite 6x^{2}-17x+12 as \left(6x^{2}-9x\right)+\left(-8x+12\right).

$$\left(6x^{2}-9x\right)+\left(-8x+12\right)$$

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

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 17x from both sides.

$$6x^{2}-17x=-12$$

Add 12 to both sides.

$$6x^{2}-17x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -17 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 6\times 12}}{2\times 6}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 6\times 12}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-17\right)±\sqrt{289-24\times 12}}{2\times 6}$$

Multiply -24 times 12.

$$x=\frac{-\left(-17\right)±\sqrt{289-288}}{2\times 6}$$

Add 289 to -288.

$$x=\frac{-\left(-17\right)±\sqrt{1}}{2\times 6}$$

Take the square root of 1.

$$x=\frac{-\left(-17\right)±1}{2\times 6}$$

The opposite of -17 is 17.

$$x=\frac{17±1}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{17±1}{12}$$

Now solve the equation x=\frac{17±1}{12} when ± is plus. Add 17 to 1.

$$x=\frac{18}{12}$$

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

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

Now solve the equation x=\frac{17±1}{12} when ± is minus. Subtract 1 from 17.

$$x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 17x from both sides.

$$6x^{2}-17x=-12$$

Divide both sides by 6.

$$\frac{6x^{2}-17x}{6}=-\frac{12}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{17}{6}x=-\frac{12}{6}$$

Divide -12 by 6.

$$x^{2}-\frac{17}{6}x=-2$$

Divide -\frac{17}{6}, the coefficient of the x term, by 2 to get -\frac{17}{12}. Then add the square of -\frac{17}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{6}x+\left(-\frac{17}{12}\right)^{2}=-2+\left(-\frac{17}{12}\right)^{2}$$

Square -\frac{17}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{6}x+\frac{289}{144}=-2+\frac{289}{144}$$

Add -2 to \frac{289}{144}.

$$x^{2}-\frac{17}{6}x+\frac{289}{144}=\frac{1}{144}$$

Factor x^{2}-\frac{17}{6}x+\frac{289}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x-\frac{17}{12}=\frac{1}{12} x-\frac{17}{12}=-\frac{1}{12}$$

Add \frac{17}{12} to both sides of the equation.

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