Solve -x^2+8=x-3

$$- x ^ { 2 } + 8 = x - 3$$

Solve for x

$$x = \frac{3 \sqrt{5} - 1}{2} \approx 2.854101966$$

$$x=\frac{-3\sqrt{5}-1}{2}\approx -3.854101966$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$-x^{2}+8-x=-3$$

Add 3 to both sides.

$$-x^{2}+8-x+3=0$$

Add 8 and 3 to get 11.

$$-x^{2}+11-x=0$$

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.

$$-x^{2}-x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 11}}{2\left(-1\right)}$$

Multiply 4 times 11.

$$x=\frac{-\left(-1\right)±\sqrt{1+44}}{2\left(-1\right)}$$

Add 1 to 44.

$$x=\frac{-\left(-1\right)±\sqrt{45}}{2\left(-1\right)}$$

Take the square root of 45.

$$x=\frac{-\left(-1\right)±3\sqrt{5}}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±3\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±3\sqrt{5}}{-2}$$

Now solve the equation x=\frac{1±3\sqrt{5}}{-2} when ± is plus. Add 1 to 3\sqrt{5}.

$$x=\frac{3\sqrt{5}+1}{-2}$$

Divide 1+3\sqrt{5} by -2.

$$x=\frac{-3\sqrt{5}-1}{2}$$

Now solve the equation x=\frac{1±3\sqrt{5}}{-2} when ± is minus. Subtract 3\sqrt{5} from 1.

$$x=\frac{1-3\sqrt{5}}{-2}$$

Divide 1-3\sqrt{5} by -2.

$$x=\frac{3\sqrt{5}-1}{2}$$

The equation is now solved.

$$x=\frac{-3\sqrt{5}-1}{2} x=\frac{3\sqrt{5}-1}{2}$$

Steps for Completing the Square

Subtract x from both sides.

$$-x^{2}+8-x=-3$$

Subtract 8 from both sides.

$$-x^{2}-x=-3-8$$

Subtract 8 from -3 to get -11.

$$-x^{2}-x=-11$$

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{11}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{11}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{11}{-1}$$

Divide -11 by -1.

$$x^{2}+x=11$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=11+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=11+\frac{1}{4}$$

Add 11 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{45}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{45}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{3\sqrt{5}}{2} x+\frac{1}{2}=-\frac{3\sqrt{5}}{2}$$

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

$$x=\frac{3\sqrt{5}-1}{2} x=\frac{-3\sqrt{5}-1}{2}$$