Solve x^2-5x+6*2-x/x^2+7x+10=2-x/x+2=

$$\frac { x ^ { 2 } - 5 x + 6 \cdot 2 - x } { x ^ { 2 } + 7 x + 10 } = \frac { 2 - x } { x + 2 } =$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{7}i}{4}\approx 0.75+0.661437828i$$

$$x=\frac{-\sqrt{7}i+3}{4}\approx 0.75-0.661437828i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,-2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x+5\right), the least common multiple of x^{2}+7x+10,x+2.

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

Multiply 6 and 2 to get 12.

$$x^{2}-5x+12-x=\left(x+5\right)\left(2-x\right)$$

Combine -5x and -x to get -6x.

$$x^{2}-6x+12=\left(x+5\right)\left(2-x\right)$$

Use the distributive property to multiply x+5 by 2-x and combine like terms.

$$x^{2}-6x+12=-3x-x^{2}+10$$

Add 3x to both sides.

$$x^{2}-6x+12+3x=-x^{2}+10$$

Combine -6x and 3x to get -3x.

$$x^{2}-3x+12=-x^{2}+10$$

Add x^{2} to both sides.

$$x^{2}-3x+12+x^{2}=10$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-3x+12=10$$

Subtract 10 from both sides.

$$2x^{2}-3x+12-10=0$$

Subtract 10 from 12 to get 2.

$$2x^{2}-3x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-16}}{2\times 2}$$

Add 9 to -16.

$$x=\frac{-\left(-3\right)±\sqrt{-7}}{2\times 2}$$

Take the square root of -7.

$$x=\frac{-\left(-3\right)±\sqrt{7}i}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{7}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±\sqrt{7}i}{4}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{4} when ± is plus. Add 3 to i\sqrt{7}.

$$x=\frac{3+\sqrt{7}i}{4}$$

Now solve the equation x=\frac{3±\sqrt{7}i}{4} when ± is minus. Subtract i\sqrt{7} from 3.

$$x=\frac{-\sqrt{7}i+3}{4}$$

The equation is now solved.

$$x=\frac{3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+3}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,-2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x+5\right), the least common multiple of x^{2}+7x+10,x+2.

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

Multiply 6 and 2 to get 12.

$$x^{2}-5x+12-x=\left(x+5\right)\left(2-x\right)$$

Combine -5x and -x to get -6x.

$$x^{2}-6x+12=\left(x+5\right)\left(2-x\right)$$

Use the distributive property to multiply x+5 by 2-x and combine like terms.

$$x^{2}-6x+12=-3x-x^{2}+10$$

Add 3x to both sides.

$$x^{2}-6x+12+3x=-x^{2}+10$$

Combine -6x and 3x to get -3x.

$$x^{2}-3x+12=-x^{2}+10$$

Add x^{2} to both sides.

$$x^{2}-3x+12+x^{2}=10$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-3x+12=10$$

Subtract 12 from both sides.

$$2x^{2}-3x=10-12$$

Subtract 12 from 10 to get -2.

$$2x^{2}-3x=-2$$

Divide both sides by 2.

$$\frac{2x^{2}-3x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{3}{2}x=-\frac{2}{2}$$

Divide -2 by 2.

$$x^{2}-\frac{3}{2}x=-1$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-1+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-1+\frac{9}{16}$$

Add -1 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{7}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{7}i}{4} x-\frac{3}{4}=-\frac{\sqrt{7}i}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+3}{4}$$