Solve x^2+x-2x-(2x^2-5)=0

$${ x }^{ 2 } +x-2x-(2 { x }^{ 2 } -5)=0$$

Solve for x

$$x = \frac{\sqrt{21} - 1}{2} \approx 1.791287847$$

$$x=\frac{-\sqrt{21}-1}{2}\approx -2.791287847$$

Steps Using the Quadratic Formula

Combine x and -2x to get -x.

$$x^{2}-x-\left(2x^{2}-5\right)=0$$

To find the opposite of 2x^{2}-5, find the opposite of each term.

$$x^{2}-x-2x^{2}+5=0$$

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

$$-x^{2}-x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 5}}{2\left(-1\right)}$$

Multiply 4 times 5.

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

Add 1 to 20.

$$x=\frac{-\left(-1\right)±\sqrt{21}}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{21}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{21}}{-2}$$

Now solve the equation x=\frac{1±\sqrt{21}}{-2} when ± is plus. Add 1 to \sqrt{21}.

$$x=\frac{\sqrt{21}+1}{-2}$$

Divide 1+\sqrt{21} by -2.

$$x=\frac{-\sqrt{21}-1}{2}$$

Now solve the equation x=\frac{1±\sqrt{21}}{-2} when ± is minus. Subtract \sqrt{21} from 1.

$$x=\frac{1-\sqrt{21}}{-2}$$

Divide 1-\sqrt{21} by -2.

$$x=\frac{\sqrt{21}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{21}-1}{2} x=\frac{\sqrt{21}-1}{2}$$

Steps for Completing the Square

Combine x and -2x to get -x.

$$x^{2}-x-\left(2x^{2}-5\right)=0$$

To find the opposite of 2x^{2}-5, find the opposite of each term.

$$x^{2}-x-2x^{2}+5=0$$

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

$$-x^{2}-x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{5}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{5}{-1}$$

Divide -5 by -1.

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

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=5+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=5+\frac{1}{4}$$

Add 5 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{21}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{21}}{2} x+\frac{1}{2}=-\frac{\sqrt{21}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{21}-1}{2} x=\frac{-\sqrt{21}-1}{2}$$