What are the solutions to the equation solvefor x,9x^2+72x-64y^2+128y+80=0 ?

The solutions to the equation solvefor x,9x^2+72x-64y^2+128y+80=0 are x=(4(2y-5))/3 ,x=-(4(2y+1))/3

Find the zeros of solvefor x,9x^2+72x-64y^2+128y+80=0

The zeros of solvefor x,9x^2+72x-64y^2+128y+80=0 are x=(4(2y-5))/3 ,x=-(4(2y+1))/3

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solve for

x, 9 x^{2} + 72 x - 64 y^{2} + 128 y + 80 = 0

x = \frac{4 ( 2 y - 5 )}{3}, x = - \frac{4 ( 2 y + 1 )}{3}

Solution steps

9 x^{2} + 72 x - 64 y^{2} + 128 y + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 72 \pm 7 2 ^{2} - 4 \cdot 9 ( - 64 y ^{2} + 128 y + 80 )}{2 \cdot 9}

Simplify

7 2^{2} - 4 \cdot 9 (- 64 y^{2} + 128 y + 80) : 48 (y - 1)

x_{1, 2} = \frac{- 72 \pm 48 ( y - 1 )}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 72 + 48 ( y - 1 )}{2 \cdot 9}, x_{2} = \frac{- 72 - 48 ( y - 1 )}{2 \cdot 9}

x = \frac{- 72 + 48 ( y - 1 )}{2 \cdot 9} : \frac{4 ( 2 y - 5 )}{3}

x = \frac{- 72 - 48 ( y - 1 )}{2 \cdot 9} : - \frac{4 ( 2 y + 1 )}{3}

The solutions to the quadratic equation are:

x = \frac{4 ( 2 y - 5 )}{3}, x = - \frac{4 ( 2 y + 1 )}{3}

(x + 3)^{2} - x = 20

x = - 4 x^{2} - 4

x^{2} + 24 = 0

a^{2} = 121

17 w^{2} = 10625

What are the solutions to the equation solvefor x,9x^2+72x-64y^2+128y+80=0 ?
The solutions to the equation solvefor x,9x^2+72x-64y^2+128y+80=0 are x=(4(2y-5))/3 ,x=-(4(2y+1))/3
Find the zeros of solvefor x,9x^2+72x-64y^2+128y+80=0
The zeros of solvefor x,9x^2+72x-64y^2+128y+80=0 are x=(4(2y-5))/3 ,x=-(4(2y+1))/3