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

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

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

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

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3 x^{2} + 4 x + c = 0

x = \frac{- 2 + - 3 c + 4}{3}, x = - \frac{- 3 c + 4 + 2}{3}

Solution steps

3 x^{2} + 4 x + c = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 3 c : 2 4 - 3 c

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- 2 + - 3 c + 4}{3}, x = - \frac{- 3 c + 4 + 2}{3}

7 x^{2} + 28 x + 56 = 0

(x + 2) (x - 2) (x^{2} - 3) = (x^{2} - 4) (x^{2} + 3)

x^{2} + 10 x + 16 = 15

y, x^{3} + x^{2} y + 3 y^{2} = 2

t, h = \frac{g t ^{2}}{2}

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