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

The solutions to the equation solvefor y,4y^2-3x=-4(x-5) are y=(sqrt(-x+20))/2 ,y=-(sqrt(-x+20))/2

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

The zeros of solvefor y,4y^2-3x=-4(x-5) are y=(sqrt(-x+20))/2 ,y=-(sqrt(-x+20))/2

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solve for

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

y = \frac{- x + 20}{2}, y = - \frac{- x + 20}{2}

Solution steps

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

Move

3 x

to the right side

4 y^{2} = - 4 (x - 5) + 3 x

Divide both sides by

4

y^{2} = \frac{- x + 20}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = \frac{- x + 20}{4}, y = - \frac{- x + 20}{4}

Simplify

\frac{- x + 20}{4} : \frac{- x + 20}{2}

Simplify

- \frac{- x + 20}{4} : - \frac{- x + 20}{2}

y = \frac{- x + 20}{2}, y = - \frac{- x + 20}{2}

(3 x + 7)^{2} - 25 = 0

3 x^{2} + 7 x - 6 = 0

y, x = 2 - y^{2}

9 n^{2} + 3 = 67

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

What are the solutions to the equation solvefor y,4y^2-3x=-4(x-5) ?
The solutions to the equation solvefor y,4y^2-3x=-4(x-5) are y=(sqrt(-x+20))/2 ,y=-(sqrt(-x+20))/2
Find the zeros of solvefor y,4y^2-3x=-4(x-5)
The zeros of solvefor y,4y^2-3x=-4(x-5) are y=(sqrt(-x+20))/2 ,y=-(sqrt(-x+20))/2