What are the solutions to the equation 3x^2-9x-24=0 ?

The solutions to the equation 3x^2-9x-24=0 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2

Find the zeros of 3x^2-9x-24=0

The zeros of 3x^2-9x-24=0 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2

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3 x^{2} - 9 x - 24 = 0

x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

Decimal Notation

x = 4.70156 \dots, x = - 1.70156 \dots

Solution steps

3 x^{2} - 9 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 ( - 24 )}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 (- 24) = 341

x_{1, 2} = \frac{- ( - 9 ) \pm 3 41}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3 41}{2 \cdot 3}, x_{2} = \frac{- ( - 9 ) - 3 41}{2 \cdot 3}

x = \frac{- ( - 9 ) + 3 41}{2 \cdot 3} : \frac{3 + 41}{2}

x = \frac{- ( - 9 ) - 3 41}{2 \cdot 3} : \frac{3 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

(125 x^{- 21} y^{6} 2^{- 15})^{- \frac{2}{3}}

x^{2} - 6 x = 11

5 x - \frac{1}{4} = 3 x - \frac{5}{4}

x + 7 - 3 > 6

\frac{7 !}{6 ! ( 7 - 6 ) !}

What are the solutions to the equation 3x^2-9x-24=0 ?
The solutions to the equation 3x^2-9x-24=0 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2
Find the zeros of 3x^2-9x-24=0
The zeros of 3x^2-9x-24=0 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2