Solve -1=5x-x^2

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

Solve for x

$$x = \frac{\sqrt{29} + 5}{2} \approx 5.192582404$$

$$x=\frac{5-\sqrt{29}}{2}\approx -0.192582404$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

Add 1 to both sides.

$$5x-x^{2}+1=0$$

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.

$$-x^{2}+5x+1=0$$

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

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4}}{2\left(-1\right)}$$

Add 25 to 4.

$$x=\frac{-5±\sqrt{29}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±\sqrt{29}}{-2}$$

Now solve the equation x=\frac{-5±\sqrt{29}}{-2} when ± is plus. Add -5 to \sqrt{29}.

$$x=\frac{\sqrt{29}-5}{-2}$$

Divide -5+\sqrt{29} by -2.

$$x=\frac{5-\sqrt{29}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{29}}{-2} when ± is minus. Subtract \sqrt{29} from -5.

$$x=\frac{-\sqrt{29}-5}{-2}$$

Divide -5-\sqrt{29} by -2.

$$x=\frac{\sqrt{29}+5}{2}$$

The equation is now solved.

$$x=\frac{5-\sqrt{29}}{2} x=\frac{\sqrt{29}+5}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 5 by -1.

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

Divide -1 by -1.

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=1+\frac{25}{4}$$

Add 1 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{29}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{29}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{29}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{29}}{2} x-\frac{5}{2}=-\frac{\sqrt{29}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{29}+5}{2} x=\frac{5-\sqrt{29}}{2}$$