What are the solutions to the equation 12x^2-48x-5=0 ?

The solutions to the equation 12x^2-48x-5=0 are x=(12+sqrt(159))/6 ,x=(12-sqrt(159))/6

Find the zeros of 12x^2-48x-5=0

The zeros of 12x^2-48x-5=0 are x=(12+sqrt(159))/6 ,x=(12-sqrt(159))/6

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12 x^{2} - 48 x - 5 = 0

x = \frac{12 + 159}{6}, x = \frac{12 - 159}{6}

Decimal Notation

x = 4.10158 \dots, x = - 0.10158 \dots

Solution steps

12 x^{2} - 48 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 12 ( - 5 )}{2 \cdot 12}

(- 48)^{2} - 4 \cdot 12 (- 5) = 4159

x_{1, 2} = \frac{- ( - 48 ) \pm 4 159}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 4 159}{2 \cdot 12}, x_{2} = \frac{- ( - 48 ) - 4 159}{2 \cdot 12}

x = \frac{- ( - 48 ) + 4 159}{2 \cdot 12} : \frac{12 + 159}{6}

x = \frac{- ( - 48 ) - 4 159}{2 \cdot 12} : \frac{12 - 159}{6}

The solutions to the quadratic equation are:

x = \frac{12 + 159}{6}, x = \frac{12 - 159}{6}

x^{2} + 4 x + 8 = 0

3 x^{2} - 48 x + 144 = 0

2 x^{2} - 4 x = 0

x^{2} + 12 x + c

x^{2} - \frac{15}{2} x + \frac{45}{4} = 0

What are the solutions to the equation 12x^2-48x-5=0 ?
The solutions to the equation 12x^2-48x-5=0 are x=(12+sqrt(159))/6 ,x=(12-sqrt(159))/6
Find the zeros of 12x^2-48x-5=0
The zeros of 12x^2-48x-5=0 are x=(12+sqrt(159))/6 ,x=(12-sqrt(159))/6