What are the solutions to the equation 12x^2-2x-3=0 ?

The solutions to the equation 12x^2-2x-3=0 are x=(1+sqrt(37))/(12),x=(1-sqrt(37))/(12)

Find the zeros of 12x^2-2x-3=0

The zeros of 12x^2-2x-3=0 are x=(1+sqrt(37))/(12),x=(1-sqrt(37))/(12)

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12 x^{2} - 2 x - 3 = 0

x = \frac{1 + 37}{12}, x = \frac{1 - 37}{12}

Decimal Notation

x = 0.59023 \dots, x = - 0.42356 \dots

Solution steps

12 x^{2} - 2 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 12 ( - 3 )}{2 \cdot 12}

(- 2)^{2} - 4 \cdot 12 (- 3) = 237

x_{1, 2} = \frac{- ( - 2 ) \pm 2 37}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 37}{2 \cdot 12}, x_{2} = \frac{- ( - 2 ) - 2 37}{2 \cdot 12}

x = \frac{- ( - 2 ) + 2 37}{2 \cdot 12} : \frac{1 + 37}{12}

x = \frac{- ( - 2 ) - 2 37}{2 \cdot 12} : \frac{1 - 37}{12}

The solutions to the quadratic equation are:

x = \frac{1 + 37}{12}, x = \frac{1 - 37}{12}

6 k + 4 < - 4 k + 6

y^{3} - 5 y^{2} - 2 y + 24

2 x = 1 + 3 x

\frac{1}{2} (x - 2) - \frac{3}{4} (x - 2) \geq \frac{1}{5} x + 1

2 - 3 p = - 3 p + 5

What are the solutions to the equation 12x^2-2x-3=0 ?
The solutions to the equation 12x^2-2x-3=0 are x=(1+sqrt(37))/(12),x=(1-sqrt(37))/(12)
Find the zeros of 12x^2-2x-3=0
The zeros of 12x^2-2x-3=0 are x=(1+sqrt(37))/(12),x=(1-sqrt(37))/(12)