What are the solutions to the equation solvefor y=3x^2-5x-4,x ?

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

Find the zeros of solvefor y=3x^2-5x-4,x

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

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solve for

y = 3 x^{2} - 5 x - 4, x

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

Solution steps

y = 3 x^{2} - 5 x - 4

Switch sides

3 x^{2} - 5 x - 4 = y

Move

y

to the left side

3 x^{2} - 5 x - 4 - y = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 r^{2} - 3 r - 8 = 0

- 5 w^{2} + 9 w + 1 = - 6 w^{2}

(x - 1)^{2} = x - 3 (x - 3 + 5)

- 4 x^{2} + 3 x + 7 = 0

x^{2} = 5 \cdot 15

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