What are the solutions to the equation solvefor x,9x^2+12xy+4y^2-48x-32y+64=0 ?

The solutions to the equation solvefor x,9x^2+12xy+4y^2-48x-32y+64=0 are x=-(2(y-4))/3

Find the zeros of solvefor x,9x^2+12xy+4y^2-48x-32y+64=0

The zeros of solvefor x,9x^2+12xy+4y^2-48x-32y+64=0 are x=-(2(y-4))/3

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solve for

x, 9 x^{2} + 12 x y + 4 y^{2} - 48 x - 32 y + 64 = 0

x = - \frac{2 ( y - 4 )}{3}

Solution steps

9 x^{2} + 12 x y + 4 y^{2} - 48 x - 32 y + 64 = 0

Write in the standard form

a x^{2} + b x + c = 0

9 x^{2} + (12 y - 48) x + 4 y^{2} - 32 y + 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 12 y - 48 ) \pm ( 12 y - 48 ) ^{2} - 4 \cdot 9 ( 4 y ^{2} - 32 y + 64 )}{2 \cdot 9}

(12 y - 48)^{2} - 4 \cdot 9 (4 y^{2} - 32 y + 64) = 0

x_{1, 2} = \frac{- ( 12 y - 48 ) \pm 0}{2 \cdot 9}

x = \frac{- ( 12 y - 48 )}{2 \cdot 9}

Simplify

\frac{- ( 12 y - 48 )}{2 \cdot 9} : - \frac{2 ( y - 4 )}{3}

x = - \frac{2 ( y - 4 )}{3}

The solution to the quadratic equation is:

x = - \frac{2 ( y - 4 )}{3}

x^{2} - 8 = - 2 x

\frac{t ^{2}}{330} - \frac{t}{5} + 24 = 0

y^{2} - y - 4 = 0

x^{2} + 12 x + 36 = 6

5 x^{2} - 4 = 16

What are the solutions to the equation solvefor x,9x^2+12xy+4y^2-48x-32y+64=0 ?
The solutions to the equation solvefor x,9x^2+12xy+4y^2-48x-32y+64=0 are x=-(2(y-4))/3
Find the zeros of solvefor x,9x^2+12xy+4y^2-48x-32y+64=0
The zeros of solvefor x,9x^2+12xy+4y^2-48x-32y+64=0 are x=-(2(y-4))/3