Solve -x^2/4-3x+4=12

$$- \frac{ { x }^{ 2 } }{ 4 } -3x+4=12$$

Solve for x

$$x=-8$$

$$x=-4$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$-x^{2}-12x+16=48$$

Subtract 48 from both sides.

$$-x^{2}-12x+16-48=0$$

Subtract 48 from 16 to get -32.

$$-x^{2}-12x-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-32. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-\left(-32\right)=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-4 b=-8$$

Rewrite -x^{2}-12x-32 as \left(-x^{2}-4x\right)+\left(-8x-32\right).

$$\left(-x^{2}-4x\right)+\left(-8x-32\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(-x-4\right)+8\left(-x-4\right)$$

Factor out common term -x-4 by using distributive property.

$$\left(-x-4\right)\left(x+8\right)$$

To find equation solutions, solve -x-4=0 and x+8=0.

$$x=-4 x=-8$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$-x^{2}-12x+16=48$$

Subtract 48 from both sides.

$$-x^{2}-12x+16-48=0$$

Subtract 48 from 16 to get -32.

$$-x^{2}-12x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -12 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-1\right)\left(-32\right)}}{2\left(-1\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-1\right)\left(-32\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-12\right)±\sqrt{144+4\left(-32\right)}}{2\left(-1\right)}$$

Multiply 4 times -32.

$$x=\frac{-\left(-12\right)±\sqrt{144-128}}{2\left(-1\right)}$$

Add 144 to -128.

$$x=\frac{-\left(-12\right)±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-\left(-12\right)±4}{2\left(-1\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{12±4}{-2}$$

Now solve the equation x=\frac{12±4}{-2} when ± is plus. Add 12 to 4.

$$x=\frac{16}{-2}$$

Divide 16 by -2.

$$x=-8$$

Now solve the equation x=\frac{12±4}{-2} when ± is minus. Subtract 4 from 12.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

The equation is now solved.

$$x=-8 x=-4$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$-x^{2}-12x+16=48$$

Subtract 16 from both sides.

$$-x^{2}-12x=48-16$$

Subtract 16 from 48 to get 32.

$$-x^{2}-12x=32$$

Divide both sides by -1.

$$\frac{-x^{2}-12x}{-1}=\frac{32}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{12}{-1}\right)x=\frac{32}{-1}$$

Divide -12 by -1.

$$x^{2}+12x=\frac{32}{-1}$$

Divide 32 by -1.

$$x^{2}+12x=-32$$

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

$$x^{2}+12x+6^{2}=-32+6^{2}$$

Square 6.

$$x^{2}+12x+36=-32+36$$

Add -32 to 36.

$$x^{2}+12x+36=4$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+6=2 x+6=-2$$

Subtract 6 from both sides of the equation.

$$x=-4 x=-8$$