Solve (-4x^2+9x+5)

$$( - 4 x ^ { 2 } + 9 x + 5 )$$

Factor

$$-4\left(x-\frac{9-\sqrt{161}}{8}\right)\left(x-\frac{\sqrt{161}+9}{8}\right)$$

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.

$$-4x^{2}+9x+5=0$$

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.

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

Square 9.

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

Multiply -4 times -4.

$$x=\frac{-9±\sqrt{81+16\times 5}}{2\left(-4\right)}$$

Multiply 16 times 5.

$$x=\frac{-9±\sqrt{81+80}}{2\left(-4\right)}$$

Add 81 to 80.

$$x=\frac{-9±\sqrt{161}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-9±\sqrt{161}}{-8}$$

Now solve the equation x=\frac{-9±\sqrt{161}}{-8} when ± is plus. Add -9 to \sqrt{161}.

$$x=\frac{\sqrt{161}-9}{-8}$$

Divide -9+\sqrt{161} by -8.

$$x=\frac{9-\sqrt{161}}{8}$$

Now solve the equation x=\frac{-9±\sqrt{161}}{-8} when ± is minus. Subtract \sqrt{161} from -9.

$$x=\frac{-\sqrt{161}-9}{-8}$$

Divide -9-\sqrt{161} by -8.

$$x=\frac{\sqrt{161}+9}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9-\sqrt{161}}{8} for x_{1} and \frac{9+\sqrt{161}}{8} for x_{2}.

$$-4x^{2}+9x+5=-4\left(x-\frac{9-\sqrt{161}}{8}\right)\left(x-\frac{\sqrt{161}+9}{8}\right)$$

Evaluate

$$5+9x-4x^{2}$$