Solve x^2-7/6x-1/2

$${ x }^{ 2 } - \frac{ 7 }{ 6 } x- \frac{ 1 }{ 2 }$$

Factor

$$\frac{\left(2x-3\right)\left(3x+1\right)}{6}$$

Solution Steps

Factor out \frac{1}{6}.

$$\frac{6x^{2}-7x-3}{6}$$

Consider 6x^{2}-7x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=6\left(-3\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -7.

$$a=-9 b=2$$

Rewrite 6x^{2}-7x-3 as \left(6x^{2}-9x\right)+\left(2x-3\right).

$$\left(6x^{2}-9x\right)+\left(2x-3\right)$$

Factor out 3x in 6x^{2}-9x.

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

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

$$\left(2x-3\right)\left(3x+1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(2x-3\right)\left(3x+1\right)}{6}$$

Evaluate

$$x^{2}-\frac{7x}{6}-\frac{1}{2}$$