Solve -(x+5/2)^2+1/4=0

$$- ( x + \frac { 5 } { 2 } ) ^ { 2 } + \frac { 1 } { 4 } = 0$$

Solve for x

$$x=-2$$

$$x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{5}{2}\right)^{2}.

$$-\left(x^{2}+5x+\frac{25}{4}\right)+\frac{1}{4}=0$$

To find the opposite of x^{2}+5x+\frac{25}{4}, find the opposite of each term.

$$-x^{2}-5x-\frac{25}{4}+\frac{1}{4}=0$$

Add -\frac{25}{4} and \frac{1}{4} to get -6.

$$-x^{2}-5x-6=0$$

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

$$a+b=-5 ab=-\left(-6\right)=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-2 b=-3$$

Rewrite -x^{2}-5x-6 as \left(-x^{2}-2x\right)+\left(-3x-6\right).

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

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

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

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

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

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

$$x=-2 x=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{5}{2}\right)^{2}.

$$-\left(x^{2}+5x+\frac{25}{4}\right)+\frac{1}{4}=0$$

To find the opposite of x^{2}+5x+\frac{25}{4}, find the opposite of each term.

$$-x^{2}-5x-\frac{25}{4}+\frac{1}{4}=0$$

Add -\frac{25}{4} and \frac{1}{4} to get -6.

$$-x^{2}-5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2\left(-1\right)}$$

Add 25 to -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-\left(-5\right)±1}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±1}{-2}$$

Now solve the equation x=\frac{5±1}{-2} when ± is plus. Add 5 to 1.

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

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{5±1}{-2} when ± is minus. Subtract 1 from 5.

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

Divide 4 by -2.

$$x=-2$$

The equation is now solved.

$$x=-3 x=-2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{5}{2}\right)^{2}.

$$-\left(x^{2}+5x+\frac{25}{4}\right)+\frac{1}{4}=0$$

To find the opposite of x^{2}+5x+\frac{25}{4}, find the opposite of each term.

$$-x^{2}-5x-\frac{25}{4}+\frac{1}{4}=0$$

Add -\frac{25}{4} and \frac{1}{4} to get -6.

$$-x^{2}-5x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$-x^{2}-5x=6$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=\frac{6}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=\frac{6}{-1}$$

Divide 6 by -1.

$$x^{2}+5x=-6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{1}{2} x+\frac{5}{2}=-\frac{1}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=-2 x=-3$$