Solve -x^2+3x+8

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

Factor

$$-\left(x-\frac{3-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+3}{2}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

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.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\times 8}}{2\left(-1\right)}$$

Multiply 4 times 8.

$$x=\frac{-3±\sqrt{9+32}}{2\left(-1\right)}$$

Add 9 to 32.

$$x=\frac{-3±\sqrt{41}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±\sqrt{41}}{-2}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{-2} when ± is plus. Add -3 to \sqrt{41}.

$$x=\frac{\sqrt{41}-3}{-2}$$

Divide -3+\sqrt{41} by -2.

$$x=\frac{3-\sqrt{41}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{-2} when ± is minus. Subtract \sqrt{41} from -3.

$$x=\frac{-\sqrt{41}-3}{-2}$$

Divide -3-\sqrt{41} by -2.

$$x=\frac{\sqrt{41}+3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3-\sqrt{41}}{2} for x_{1} and \frac{3+\sqrt{41}}{2} for x_{2}.

$$-x^{2}+3x+8=-\left(x-\frac{3-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+3}{2}\right)$$

Evaluate

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