Solve 7x=x^2-1

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

Solve for x

$$x = \frac{\sqrt{53} + 7}{2} \approx 7.140054945$$

$$x=\frac{7-\sqrt{53}}{2}\approx -0.140054945$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Add 1 to both sides.

$$7x-x^{2}+1=0$$

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.

$$-x^{2}+7x+1=0$$

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

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

Square 7.

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

Multiply -4 times -1.

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

Add 49 to 4.

$$x=\frac{-7±\sqrt{53}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±\sqrt{53}}{-2}$$

Now solve the equation x=\frac{-7±\sqrt{53}}{-2} when ± is plus. Add -7 to \sqrt{53}.

$$x=\frac{\sqrt{53}-7}{-2}$$

Divide -7+\sqrt{53} by -2.

$$x=\frac{7-\sqrt{53}}{2}$$

Now solve the equation x=\frac{-7±\sqrt{53}}{-2} when ± is minus. Subtract \sqrt{53} from -7.

$$x=\frac{-\sqrt{53}-7}{-2}$$

Divide -7-\sqrt{53} by -2.

$$x=\frac{\sqrt{53}+7}{2}$$

The equation is now solved.

$$x=\frac{7-\sqrt{53}}{2} x=\frac{\sqrt{53}+7}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 7 by -1.

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

Divide -1 by -1.

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

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.

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

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

$$x^{2}-7x+\frac{49}{4}=1+\frac{49}{4}$$

Add 1 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{53}{4}$$

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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{53}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{53}}{2} x-\frac{7}{2}=-\frac{\sqrt{53}}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{53}+7}{2} x=\frac{7-\sqrt{53}}{2}$$