What are the solutions to the equation 24x^2-16x-39=0 ?

The solutions to the equation 24x^2-16x-39=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)

Find the zeros of 24x^2-16x-39=0

The zeros of 24x^2-16x-39=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)

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24 x^{2} - 16 x - 39 = 0

x = \frac{4 + 5 10}{12}, x = \frac{4 - 5 10}{12}

Decimal Notation

x = 1.65094 \dots, x = - 0.98428 \dots

Solution steps

24 x^{2} - 16 x - 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 24 ( - 39 )}{2 \cdot 24}

(- 16)^{2} - 4 \cdot 24 (- 39) = 2010

x_{1, 2} = \frac{- ( - 16 ) \pm 20 10}{2 \cdot 24}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 20 10}{2 \cdot 24}, x_{2} = \frac{- ( - 16 ) - 20 10}{2 \cdot 24}

x = \frac{- ( - 16 ) + 20 10}{2 \cdot 24} : \frac{4 + 5 10}{12}

x = \frac{- ( - 16 ) - 20 10}{2 \cdot 24} : \frac{4 - 5 10}{12}

The solutions to the quadratic equation are:

x = \frac{4 + 5 10}{12}, x = \frac{4 - 5 10}{12}

\frac{1}{4} + 7

(2 x - 1) (4 x - 3) = 0

\frac{x}{y}

5 (2 + v) - v = 10 + 4 (v + 1)

6 x^{3} + 13 x^{2} - 4

What are the solutions to the equation 24x^2-16x-39=0 ?
The solutions to the equation 24x^2-16x-39=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)
Find the zeros of 24x^2-16x-39=0
The zeros of 24x^2-16x-39=0 are x=(4+5sqrt(10))/(12),x=(4-5sqrt(10))/(12)