What are the solutions to the equation 3x^2+5x-4+n=0 ?

The solutions to the equation 3x^2+5x-4+n=0 are x=(-5+sqrt(-12n+73))/6 ,x=(-5-sqrt(-12n+73))/6

Find the zeros of 3x^2+5x-4+n=0

The zeros of 3x^2+5x-4+n=0 are x=(-5+sqrt(-12n+73))/6 ,x=(-5-sqrt(-12n+73))/6

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3 x^{2} + 5 x - 4 + n = 0

x = \frac{- 5 + - 12 n + 73}{6}, x = \frac{- 5 - - 12 n + 73}{6}

Solution steps

3 x^{2} + 5 x - 4 + n = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 ( - 4 + n )}{2 \cdot 3}

Simplify

5^{2} - 4 \cdot 3 (- 4 + n) : - 12 n + 73

x_{1, 2} = \frac{- 5 \pm - 12 n + 73}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 5 + - 12 n + 73}{2 \cdot 3}, x_{2} = \frac{- 5 - - 12 n + 73}{2 \cdot 3}

x = \frac{- 5 + - 12 n + 73}{2 \cdot 3} : \frac{- 5 + - 12 n + 73}{6}

x = \frac{- 5 - - 12 n + 73}{2 \cdot 3} : \frac{- 5 - - 12 n + 73}{6}

The solutions to the quadratic equation are:

x = \frac{- 5 + - 12 n + 73}{6}, x = \frac{- 5 - - 12 n + 73}{6}

x^{2} - 10 x - 400 = 0

11 x^{2} + 12 = 0

1.5 t^{2} - 17.6 t - 75 = 0

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

3 (x - 2) = 2 (x^{2} - 9) + x^{2} - 5 (x + 2)

What are the solutions to the equation 3x^2+5x-4+n=0 ?
The solutions to the equation 3x^2+5x-4+n=0 are x=(-5+sqrt(-12n+73))/6 ,x=(-5-sqrt(-12n+73))/6
Find the zeros of 3x^2+5x-4+n=0
The zeros of 3x^2+5x-4+n=0 are x=(-5+sqrt(-12n+73))/6 ,x=(-5-sqrt(-12n+73))/6