Solve 9x^2+8x-9

$$9 { x }^{ 2 } +8x-9$$

Factor

$$9\left(x-\frac{-\sqrt{97}-4}{9}\right)\left(x-\frac{\sqrt{97}-4}{9}\right)$$

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.

$$9x^{2}+8x-9=0$$

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.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 9\left(-9\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-8±\sqrt{64-36\left(-9\right)}}{2\times 9}$$

Multiply -36 times -9.

$$x=\frac{-8±\sqrt{64+324}}{2\times 9}$$

Add 64 to 324.

$$x=\frac{-8±\sqrt{388}}{2\times 9}$$

Take the square root of 388.

$$x=\frac{-8±2\sqrt{97}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-8±2\sqrt{97}}{18}$$

Now solve the equation x=\frac{-8±2\sqrt{97}}{18} when ± is plus. Add -8 to 2\sqrt{97}.

$$x=\frac{2\sqrt{97}-8}{18}$$

Divide -8+2\sqrt{97} by 18.

$$x=\frac{\sqrt{97}-4}{9}$$

Now solve the equation x=\frac{-8±2\sqrt{97}}{18} when ± is minus. Subtract 2\sqrt{97} from -8.

$$x=\frac{-2\sqrt{97}-8}{18}$$

Divide -8-2\sqrt{97} by 18.

$$x=\frac{-\sqrt{97}-4}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-4+\sqrt{97}}{9} for x_{1} and \frac{-4-\sqrt{97}}{9} for x_{2}.

$$9x^{2}+8x-9=9\left(x-\frac{\sqrt{97}-4}{9}\right)\left(x-\frac{-\sqrt{97}-4}{9}\right)$$

Evaluate

$$9x^{2}+8x-9$$