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

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

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

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

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3 x^{2} + 4 x = 12

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

Move

12

to the left side

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}

2 x^{2} - 3 = x + 3

x^{2} + 8^{2} = 9^{2}

m^{2} + 2 m - 1 = 0

e^{2} = 2

3 n^{2} + 7 n - 20 = 0

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