What are the solutions to the equation -3x^2+11x+9=1 ?

The solutions to the equation -3x^2+11x+9=1 are x=-(-11+sqrt(217))/6 ,x=(11+sqrt(217))/6

Find the zeros of -3x^2+11x+9=1

The zeros of -3x^2+11x+9=1 are x=-(-11+sqrt(217))/6 ,x=(11+sqrt(217))/6

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- 3 x^{2} + 11 x + 9 = 1

x = - \frac{- 11 + 217}{6}, x = \frac{11 + 217}{6}

Decimal Notation

x = - 0.62181 \dots, x = 4.28848 \dots

Solution steps

- 3 x^{2} + 11 x + 9 = 1

Move

1

to the left side

- 3 x^{2} + 11 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 3 ) \cdot 8}{2 ( - 3 )}

1 1^{2} - 4 (- 3) \cdot 8 = 217

x_{1, 2} = \frac{- 11 \pm 217}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 11 + 217}{2 ( - 3 )}, x_{2} = \frac{- 11 - 217}{2 ( - 3 )}

x = \frac{- 11 + 217}{2 ( - 3 )} : - \frac{- 11 + 217}{6}

x = \frac{- 11 - 217}{2 ( - 3 )} : \frac{11 + 217}{6}

The solutions to the quadratic equation are:

x = - \frac{- 11 + 217}{6}, x = \frac{11 + 217}{6}

5.56 \cdot 1 0^{- 10} = \frac{x ^{2}}{2.5}

x^{2} - 1 = - 2 x

x, x^{2} = \frac{1}{y ^{3}}

(x + 1)^{2} = x + 1

0.2 x^{2} - 1.5 x + 2.9 = 0

What are the solutions to the equation -3x^2+11x+9=1 ?
The solutions to the equation -3x^2+11x+9=1 are x=-(-11+sqrt(217))/6 ,x=(11+sqrt(217))/6
Find the zeros of -3x^2+11x+9=1
The zeros of -3x^2+11x+9=1 are x=-(-11+sqrt(217))/6 ,x=(11+sqrt(217))/6