Solve (-2x-1)-4x+6=2x^2+9-8x

$$( - 2 x - 1 ) - 4 x + 6 = 2 x ^ { 2 } + 9 - 8 x$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{7}i}{2}\approx 0.5+1.322875656i$$

$$x=\frac{-\sqrt{7}i+1}{2}\approx 0.5-1.322875656i$$

Steps Using the Quadratic Formula

Combine -2x and -4x to get -6x.

$$-6x-1+6=2x^{2}+9-8x$$

Add -1 and 6 to get 5.

$$-6x+5=2x^{2}+9-8x$$

Subtract 2x^{2} from both sides.

$$-6x+5-2x^{2}=9-8x$$

Subtract 9 from both sides.

$$-6x+5-2x^{2}-9=-8x$$

Subtract 9 from 5 to get -4.

$$-6x-4-2x^{2}=-8x$$

Add 8x to both sides.

$$-6x-4-2x^{2}+8x=0$$

Combine -6x and 8x to get 2x.

$$2x-4-2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+2x-4=0$$

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

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

Square 2.

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

Multiply -4 times -2.

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

Multiply 8 times -4.

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

Add 4 to -32.

$$x=\frac{-2±\sqrt{-28}}{2\left(-2\right)}$$

Take the square root of -28.

$$x=\frac{-2±2\sqrt{7}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±2\sqrt{7}i}{-4}$$

Now solve the equation x=\frac{-2±2\sqrt{7}i}{-4} when ± is plus. Add -2 to 2i\sqrt{7}.

$$x=\frac{-2+2\sqrt{7}i}{-4}$$

Divide -2+2i\sqrt{7} by -4.

$$x=\frac{-\sqrt{7}i+1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{7}i}{-4} when ± is minus. Subtract 2i\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}i-2}{-4}$$

Divide -2-2i\sqrt{7} by -4.

$$x=\frac{1+\sqrt{7}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}i+1}{2} x=\frac{1+\sqrt{7}i}{2}$$

Steps for Completing the Square

Combine -2x and -4x to get -6x.

$$-6x-1+6=2x^{2}+9-8x$$

Add -1 and 6 to get 5.

$$-6x+5=2x^{2}+9-8x$$

Subtract 2x^{2} from both sides.

$$-6x+5-2x^{2}=9-8x$$

Add 8x to both sides.

$$-6x+5-2x^{2}+8x=9$$

Combine -6x and 8x to get 2x.

$$2x+5-2x^{2}=9$$

Subtract 5 from both sides.

$$2x-2x^{2}=9-5$$

Subtract 5 from 9 to get 4.

$$2x-2x^{2}=4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+2x=4$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 2 by -2.

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

Divide 4 by -2.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{7}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{7}i}{2} x-\frac{1}{2}=-\frac{\sqrt{7}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{1+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i+1}{2}$$