What are the solutions to the equation solvefor x,2y^2-3xy-2x^2+10x+5y-12=0 ?

The solutions to the equation solvefor x,2y^2-3xy-2x^2+10x+5y-12=0 are x=-2y+3,x=(y+4)/2

Find the zeros of solvefor x,2y^2-3xy-2x^2+10x+5y-12=0

The zeros of solvefor x,2y^2-3xy-2x^2+10x+5y-12=0 are x=-2y+3,x=(y+4)/2

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solve for

x, 2 y^{2} - 3 x y - 2 x^{2} + 10 x + 5 y - 12 = 0

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

Solution steps

2 y^{2} - 3 x y - 2 x^{2} + 10 x + 5 y - 12 = 0

Write in the standard form

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

- 2 x^{2} + (- 3 y + 10) x + 2 y^{2} + 5 y - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 y + 10 ) \pm ( - 3 y + 10 ) ^{2} - 4 ( - 2 ) ( 2 y ^{2} + 5 y - 12 )}{2 ( - 2 )}

Simplify

(- 3 y + 10)^{2} - 4 (- 2) (2 y^{2} + 5 y - 12) : 5 y - 2

x_{1, 2} = \frac{- ( - 3 y + 10 ) \pm ( 5 y - 2 )}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 y + 10 ) + 5 y - 2}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 y + 10 ) - ( 5 y - 2 )}{2 ( - 2 )}

x = \frac{- ( - 3 y + 10 ) + 5 y - 2}{2 ( - 2 )} : - 2 y + 3

x = \frac{- ( - 3 y + 10 ) - ( 5 y - 2 )}{2 ( - 2 )} : \frac{y + 4}{2}

The solutions to the quadratic equation are:

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

140 = 110 (1 + r)^{2}

x, 12 x^{2} - 26 x y + 12 y^{2} = 0

y, y^{2} = x^{2} (1 - x)

z^{1} + z^{2} - 2 = 0

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

What are the solutions to the equation solvefor x,2y^2-3xy-2x^2+10x+5y-12=0 ?
The solutions to the equation solvefor x,2y^2-3xy-2x^2+10x+5y-12=0 are x=-2y+3,x=(y+4)/2
Find the zeros of solvefor x,2y^2-3xy-2x^2+10x+5y-12=0
The zeros of solvefor x,2y^2-3xy-2x^2+10x+5y-12=0 are x=-2y+3,x=(y+4)/2