What are the solutions to the equation 0=12x^2-24x-40 ?

The solutions to the equation 0=12x^2-24x-40 are x=(3+sqrt(39))/3 ,x=(3-sqrt(39))/3

Find the zeros of 0=12x^2-24x-40

The zeros of 0=12x^2-24x-40 are x=(3+sqrt(39))/3 ,x=(3-sqrt(39))/3

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0 = 12 x^{2} - 24 x - 40

x = \frac{3 + 39}{3}, x = \frac{3 - 39}{3}

Decimal Notation

x = 3.08166 \dots, x = - 1.08166 \dots

Solution steps

0 = 12 x^{2} - 24 x - 40

Switch sides

12 x^{2} - 24 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 12 ( - 40 )}{2 \cdot 12}

(- 24)^{2} - 4 \cdot 12 (- 40) = 839

x_{1, 2} = \frac{- ( - 24 ) \pm 8 39}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 8 39}{2 \cdot 12}, x_{2} = \frac{- ( - 24 ) - 8 39}{2 \cdot 12}

x = \frac{- ( - 24 ) + 8 39}{2 \cdot 12} : \frac{3 + 39}{3}

x = \frac{- ( - 24 ) - 8 39}{2 \cdot 12} : \frac{3 - 39}{3}

The solutions to the quadratic equation are:

x = \frac{3 + 39}{3}, x = \frac{3 - 39}{3}

- 6 x^{2} + 72 x - 192 = 0

2 x^{2} - 16 x + 31

x^{2} + 18 x + 17 = 0

- 0.3 x^{2} + 150 x + 70 = 0

r^{2} - 3 r - 28 = 0

What are the solutions to the equation 0=12x^2-24x-40 ?
The solutions to the equation 0=12x^2-24x-40 are x=(3+sqrt(39))/3 ,x=(3-sqrt(39))/3
Find the zeros of 0=12x^2-24x-40
The zeros of 0=12x^2-24x-40 are x=(3+sqrt(39))/3 ,x=(3-sqrt(39))/3