What are the solutions to the equation 9x^2+6x-32=0 ?

The solutions to the equation 9x^2+6x-32=0 are x=(-1+sqrt(33))/3 ,x=-(1+sqrt(33))/3

Find the zeros of 9x^2+6x-32=0

The zeros of 9x^2+6x-32=0 are x=(-1+sqrt(33))/3 ,x=-(1+sqrt(33))/3

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9 x^{2} + 6 x - 32 = 0

x = \frac{- 1 + 33}{3}, x = - \frac{1 + 33}{3}

Decimal Notation

x = 1.58152 \dots, x = - 2.24818 \dots

Solution steps

9 x^{2} + 6 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 ( - 32 )}{2 \cdot 9}

6^{2} - 4 \cdot 9 (- 32) = 633

x_{1, 2} = \frac{- 6 \pm 6 33}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 6 + 6 33}{2 \cdot 9}, x_{2} = \frac{- 6 - 6 33}{2 \cdot 9}

x = \frac{- 6 + 6 33}{2 \cdot 9} : \frac{- 1 + 33}{3}

x = \frac{- 6 - 6 33}{2 \cdot 9} : - \frac{1 + 33}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 33}{3}, x = - \frac{1 + 33}{3}

8 P^{2} + 12 P - 8

7 x^{2} + 3 x

- x = 3

\frac{1}{2} \geq - \frac{1}{6} - \frac{2}{9} x

- 7 x + 3 \leq 4 - x

What are the solutions to the equation 9x^2+6x-32=0 ?
The solutions to the equation 9x^2+6x-32=0 are x=(-1+sqrt(33))/3 ,x=-(1+sqrt(33))/3
Find the zeros of 9x^2+6x-32=0
The zeros of 9x^2+6x-32=0 are x=(-1+sqrt(33))/3 ,x=-(1+sqrt(33))/3