Solve -x^2/4-3x+4=9

$$- \frac{ { x }^{ 2 } }{ 4 } -3x+4 = 9$$

Solve for x

$$x=-10$$

$$x=-2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$-x^{2}-12x+16=36$$

Subtract 36 from both sides.

$$-x^{2}-12x+16-36=0$$

Subtract 36 from 16 to get -20.

$$-x^{2}-12x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-\left(-20\right)=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$a=-2 b=-10$$

Rewrite -x^{2}-12x-20 as \left(-x^{2}-2x\right)+\left(-10x-20\right).

$$\left(-x^{2}-2x\right)+\left(-10x-20\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(-x-2\right)+10\left(-x-2\right)$$

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

$$\left(-x-2\right)\left(x+10\right)$$

To find equation solutions, solve -x-2=0 and x+10=0.

$$x=-2 x=-10$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$-x^{2}-12x+16=36$$

Subtract 36 from both sides.

$$-x^{2}-12x+16-36=0$$

Subtract 36 from 16 to get -20.

$$-x^{2}-12x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -12 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-12\right)±\sqrt{144+4\left(-20\right)}}{2\left(-1\right)}$$

Multiply 4 times -20.

$$x=\frac{-\left(-12\right)±\sqrt{144-80}}{2\left(-1\right)}$$

Add 144 to -80.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{-\left(-12\right)±8}{2\left(-1\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{12±8}{-2}$$

Now solve the equation x=\frac{12±8}{-2} when ± is plus. Add 12 to 8.

$$x=\frac{20}{-2}$$

Divide 20 by -2.

$$x=-10$$

Now solve the equation x=\frac{12±8}{-2} when ± is minus. Subtract 8 from 12.

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

Divide 4 by -2.

$$x=-2$$

The equation is now solved.

$$x=-10 x=-2$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$-x^{2}-12x+16=36$$

Subtract 16 from both sides.

$$-x^{2}-12x=36-16$$

Subtract 16 from 36 to get 20.

$$-x^{2}-12x=20$$

Divide both sides by -1.

$$\frac{-x^{2}-12x}{-1}=\frac{20}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{12}{-1}\right)x=\frac{20}{-1}$$

Divide -12 by -1.

$$x^{2}+12x=\frac{20}{-1}$$

Divide 20 by -1.

$$x^{2}+12x=-20$$

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.

$$x^{2}+12x+6^{2}=-20+6^{2}$$

Square 6.

$$x^{2}+12x+36=-20+36$$

Add -20 to 36.

$$x^{2}+12x+36=16$$

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}.

$$\left(x+6\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+6=4 x+6=-4$$

Subtract 6 from both sides of the equation.

$$x=-2 x=-10$$