Solve 2-x^2=7x

$$2 - x ^ { 2 } = 7 x$$

Solve for x

$$x=\frac{\sqrt{57}-7}{2}\approx 0.274917218$$

$$x=\frac{-\sqrt{57}-7}{2}\approx -7.274917218$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$2-x^{2}-7x=0$$

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.

$$-x^{2}-7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49+8}}{2\left(-1\right)}$$

Add 49 to 8.

$$x=\frac{-\left(-7\right)±\sqrt{57}}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{57}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{7±\sqrt{57}}{-2}$$

Now solve the equation x=\frac{7±\sqrt{57}}{-2} when ± is plus. Add 7 to \sqrt{57}.

$$x=\frac{\sqrt{57}+7}{-2}$$

Divide 7+\sqrt{57} by -2.

$$x=\frac{-\sqrt{57}-7}{2}$$

Now solve the equation x=\frac{7±\sqrt{57}}{-2} when ± is minus. Subtract \sqrt{57} from 7.

$$x=\frac{7-\sqrt{57}}{-2}$$

Divide 7-\sqrt{57} by -2.

$$x=\frac{\sqrt{57}-7}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{57}-7}{2} x=\frac{\sqrt{57}-7}{2}$$

Steps for Completing the Square

Subtract 7x from both sides.

$$2-x^{2}-7x=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}-7x=-2$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -7 by -1.

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

Divide -2 by -1.

$$x^{2}+7x=2$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=2+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=2+\frac{49}{4}$$

Add 2 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{57}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{57}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{57}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{57}}{2} x+\frac{7}{2}=-\frac{\sqrt{57}}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=\frac{\sqrt{57}-7}{2} x=\frac{-\sqrt{57}-7}{2}$$