Solve -x^2+1/2x+3

$$- x ^ { 2 } + \frac { 1 } { 2 } x + 3$$

Factor

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

Solution Steps

Factor out \frac{1}{2}.

$$\frac{-2x^{2}+x+6}{2}$$

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

$$a+b=1 ab=-2\times 6=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -2x^{2}+x+6 as \left(-2x^{2}+4x\right)+\left(-3x+6\right).

$$\left(-2x^{2}+4x\right)+\left(-3x+6\right)$$

Factor out 2x in the first and 3 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

$$-x^{2}+\frac{x}{2}+3$$