What are the solutions to the equation 3x^2+4x-12=0 ?

The solutions to the equation 3x^2+4x-12=0 are x=(2(sqrt(10)-1))/3 ,x=-(2(1+sqrt(10)))/3

Find the zeros of 3x^2+4x-12=0

The zeros of 3x^2+4x-12=0 are x=(2(sqrt(10)-1))/3 ,x=-(2(1+sqrt(10)))/3

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3 x^{2} + 4 x - 12 = 0

x = \frac{2 ( 10 - 1 )}{3}, x = - \frac{2 ( 1 + 10 )}{3}

Decimal Notation

x = 1.44151 \dots, x = - 2.77485 \dots

Solution steps

3 x^{2} + 4 x - 12 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 3 (- 12) = 410

x_{1, 2} = \frac{- 4 \pm 4 10}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 4 + 4 10}{2 \cdot 3}, x_{2} = \frac{- 4 - 4 10}{2 \cdot 3}

x = \frac{- 4 + 4 10}{2 \cdot 3} : \frac{2 ( 10 - 1 )}{3}

x = \frac{- 4 - 4 10}{2 \cdot 3} : - \frac{2 ( 1 + 10 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 10 - 1 )}{3}, x = - \frac{2 ( 1 + 10 )}{3}

x, x^{2} - 2 x = 0

x^{2} + 9 x - 18 = 0

12 x^{2} + 4 x = 0

12 x^{2} - 252 x + 902 = 0

7 m^{2} + 40 = 43 m

What are the solutions to the equation 3x^2+4x-12=0 ?
The solutions to the equation 3x^2+4x-12=0 are x=(2(sqrt(10)-1))/3 ,x=-(2(1+sqrt(10)))/3
Find the zeros of 3x^2+4x-12=0
The zeros of 3x^2+4x-12=0 are x=(2(sqrt(10)-1))/3 ,x=-(2(1+sqrt(10)))/3