Solve -5+9x-2x^2

$$- 5 + 9 x - 2 x ^ { 2 }$$

Factor

$$-2\left(x-\frac{9-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+9}{4}\right)$$

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.

$$-2x^{2}+9x-5=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-9±\sqrt{81+8\left(-5\right)}}{2\left(-2\right)}$$

Multiply 8 times -5.

$$x=\frac{-9±\sqrt{81-40}}{2\left(-2\right)}$$

Add 81 to -40.

$$x=\frac{-9±\sqrt{41}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-9±\sqrt{41}}{-4}$$

Now solve the equation x=\frac{-9±\sqrt{41}}{-4} when ± is plus. Add -9 to \sqrt{41}.

$$x=\frac{\sqrt{41}-9}{-4}$$

Divide -9+\sqrt{41} by -4.

$$x=\frac{9-\sqrt{41}}{4}$$

Now solve the equation x=\frac{-9±\sqrt{41}}{-4} when ± is minus. Subtract \sqrt{41} from -9.

$$x=\frac{-\sqrt{41}-9}{-4}$$

Divide -9-\sqrt{41} by -4.

$$x=\frac{\sqrt{41}+9}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9-\sqrt{41}}{4} for x_{1} and \frac{9+\sqrt{41}}{4} for x_{2}.

$$-2x^{2}+9x-5=-2\left(x-\frac{9-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+9}{4}\right)$$

Evaluate

$$-2x^{2}+9x-5$$