Solve -x^2-9x+12=-x-6

$$- x ^ { 2 } - 9 x + 12 = - x - 6$$

Solve for x (complex solution)

$$x=\sqrt{34}-4\approx 1.830951895$$

$$x=-\left(\sqrt{34}+4\right)\approx -9.830951895$$

Steps Using the Quadratic Formula

Add x to both sides.

$$-x^{2}-9x+12+x=-6$$

Combine -9x and x to get -8x.

$$-x^{2}-8x+12=-6$$

Add 6 to both sides.

$$-x^{2}-8x+12+6=0$$

Add 12 and 6 to get 18.

$$-x^{2}-8x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\times 18}}{2\left(-1\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\times 18}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-8\right)±\sqrt{64+4\times 18}}{2\left(-1\right)}$$

Multiply 4 times 18.

$$x=\frac{-\left(-8\right)±\sqrt{64+72}}{2\left(-1\right)}$$

Add 64 to 72.

$$x=\frac{-\left(-8\right)±\sqrt{136}}{2\left(-1\right)}$$

Take the square root of 136.

$$x=\frac{-\left(-8\right)±2\sqrt{34}}{2\left(-1\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{34}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{8±2\sqrt{34}}{-2}$$

Now solve the equation x=\frac{8±2\sqrt{34}}{-2} when ± is plus. Add 8 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}+8}{-2}$$

Divide 8+2\sqrt{34} by -2.

$$x=-\left(\sqrt{34}+4\right)$$

Now solve the equation x=\frac{8±2\sqrt{34}}{-2} when ± is minus. Subtract 2\sqrt{34} from 8.

$$x=\frac{8-2\sqrt{34}}{-2}$$

Divide 8-2\sqrt{34} by -2.

$$x=\sqrt{34}-4$$

The equation is now solved.

$$x=-\left(\sqrt{34}+4\right) x=\sqrt{34}-4$$

Steps for Completing the Square

Add x to both sides.

$$-x^{2}-9x+12+x=-6$$

Combine -9x and x to get -8x.

$$-x^{2}-8x+12=-6$$

Subtract 12 from both sides.

$$-x^{2}-8x=-6-12$$

Subtract 12 from -6 to get -18.

$$-x^{2}-8x=-18$$

Divide both sides by -1.

$$\frac{-x^{2}-8x}{-1}=-\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{8}{-1}\right)x=-\frac{18}{-1}$$

Divide -8 by -1.

$$x^{2}+8x=-\frac{18}{-1}$$

Divide -18 by -1.

$$x^{2}+8x=18$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=18+4^{2}$$

Square 4.

$$x^{2}+8x+16=18+16$$

Add 18 to 16.

$$x^{2}+8x+16=34$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=34$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{34}$$

Simplify.

$$x+4=\sqrt{34} x+4=-\sqrt{34}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{34}-4 x=-\sqrt{34}-4$$

Solve for x

$$x=\sqrt{34}-4\approx 1.830951895$$

$$x=-\sqrt{34}-4\approx -9.830951895$$

Steps Using the Quadratic Formula

Add x to both sides.

$$-x^{2}-9x+12+x=-6$$

Combine -9x and x to get -8x.

$$-x^{2}-8x+12=-6$$

Add 6 to both sides.

$$-x^{2}-8x+12+6=0$$

Add 12 and 6 to get 18.

$$-x^{2}-8x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\times 18}}{2\left(-1\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\times 18}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-8\right)±\sqrt{64+4\times 18}}{2\left(-1\right)}$$

Multiply 4 times 18.

$$x=\frac{-\left(-8\right)±\sqrt{64+72}}{2\left(-1\right)}$$

Add 64 to 72.

$$x=\frac{-\left(-8\right)±\sqrt{136}}{2\left(-1\right)}$$

Take the square root of 136.

$$x=\frac{-\left(-8\right)±2\sqrt{34}}{2\left(-1\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{34}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{8±2\sqrt{34}}{-2}$$

Now solve the equation x=\frac{8±2\sqrt{34}}{-2} when ± is plus. Add 8 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}+8}{-2}$$

Divide 8+2\sqrt{34} by -2.

$$x=-\left(\sqrt{34}+4\right)$$

Now solve the equation x=\frac{8±2\sqrt{34}}{-2} when ± is minus. Subtract 2\sqrt{34} from 8.

$$x=\frac{8-2\sqrt{34}}{-2}$$

Divide 8-2\sqrt{34} by -2.

$$x=\sqrt{34}-4$$

The equation is now solved.

$$x=-\left(\sqrt{34}+4\right) x=\sqrt{34}-4$$

Steps for Completing the Square

Add x to both sides.

$$-x^{2}-9x+12+x=-6$$

Combine -9x and x to get -8x.

$$-x^{2}-8x+12=-6$$

Subtract 12 from both sides.

$$-x^{2}-8x=-6-12$$

Subtract 12 from -6 to get -18.

$$-x^{2}-8x=-18$$

Divide both sides by -1.

$$\frac{-x^{2}-8x}{-1}=-\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{8}{-1}\right)x=-\frac{18}{-1}$$

Divide -8 by -1.

$$x^{2}+8x=-\frac{18}{-1}$$

Divide -18 by -1.

$$x^{2}+8x=18$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=18+4^{2}$$

Square 4.

$$x^{2}+8x+16=18+16$$

Add 18 to 16.

$$x^{2}+8x+16=34$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=34$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{34}$$

Simplify.

$$x+4=\sqrt{34} x+4=-\sqrt{34}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{34}-4 x=-\sqrt{34}-4$$