What are the solutions to the equation 3x^2+4x-12=13 ?

The solutions to the equation 3x^2+4x-12=13 are x=(-2+sqrt(79))/3 ,x=-(2+sqrt(79))/3

Find the zeros of 3x^2+4x-12=13

The zeros of 3x^2+4x-12=13 are x=(-2+sqrt(79))/3 ,x=-(2+sqrt(79))/3

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3 x^{2} + 4 x - 12 = 13

x = \frac{- 2 + 79}{3}, x = - \frac{2 + 79}{3}

Decimal Notation

x = 2.29606 \dots, x = - 3.62939 \dots

Solution steps

3 x^{2} + 4 x - 12 = 13

Move

13

to the left side

3 x^{2} + 4 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 ( - 25 )}{2 \cdot 3}

4^{2} - 4 \cdot 3 (- 25) = 279

x_{1, 2} = \frac{- 4 \pm 2 79}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 4 + 2 79}{2 \cdot 3}, x_{2} = \frac{- 4 - 2 79}{2 \cdot 3}

x = \frac{- 4 + 2 79}{2 \cdot 3} : \frac{- 2 + 79}{3}

x = \frac{- 4 - 2 79}{2 \cdot 3} : - \frac{2 + 79}{3}

The solutions to the quadratic equation are:

x = \frac{- 2 + 79}{3}, x = - \frac{2 + 79}{3}

x^{2} - 5 x = 21

h, (2 - t) ((5 - h - t) (5 + h - t)) + h^{2} = 0

x^{2} - x + 1 = 2 x - 1

x^{2} - 5 x = - 7

(37.15 - 1 x) \cdot (10.19 - 0.5 x) - (15.12)^{2} = 0

What are the solutions to the equation 3x^2+4x-12=13 ?
The solutions to the equation 3x^2+4x-12=13 are x=(-2+sqrt(79))/3 ,x=-(2+sqrt(79))/3
Find the zeros of 3x^2+4x-12=13
The zeros of 3x^2+4x-12=13 are x=(-2+sqrt(79))/3 ,x=-(2+sqrt(79))/3