What are the solutions to the equation solvefor y,y^2-4x-6y+1=0 ?

The solutions to the equation solvefor y,y^2-4x-6y+1=0 are y=3+2sqrt(x+2),y=3-2sqrt(x+2)

Find the zeros of solvefor y,y^2-4x-6y+1=0

The zeros of solvefor y,y^2-4x-6y+1=0 are y=3+2sqrt(x+2),y=3-2sqrt(x+2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

y, y^{2} - 4 x - 6 y + 1 = 0

y = 3 + 2 x + 2, y = 3 - 2 x + 2

Solution steps

y^{2} - 4 x - 6 y + 1 = 0

Write in the standard form

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

y^{2} - 6 y - 4 x + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 4 x + 1 )}{2 \cdot 1}

Simplify

(- 6)^{2} - 4 \cdot 1 \cdot (- 4 x + 1) : 4 x + 2

y_{1, 2} = \frac{- ( - 6 ) \pm 4 x + 2}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 6 ) + 4 x + 2}{2 \cdot 1}, y_{2} = \frac{- ( - 6 ) - 4 x + 2}{2 \cdot 1}

y = \frac{- ( - 6 ) + 4 x + 2}{2 \cdot 1} : 3 + 2 x + 2

y = \frac{- ( - 6 ) - 4 x + 2}{2 \cdot 1} : 3 - 2 x + 2

The solutions to the quadratic equation are:

y = 3 + 2 x + 2, y = 3 - 2 x + 2

(3 x + 3) (x - 1) = x (x + 1)

x^{2} - 4 x = 6 x - x^{2}

x^{2} = 1 5^{2} + 3 6^{2}

8 x^{2} - 2 = x^{2} + 12

d, v = \frac{π d ^{2} h}{4}

What are the solutions to the equation solvefor y,y^2-4x-6y+1=0 ?
The solutions to the equation solvefor y,y^2-4x-6y+1=0 are y=3+2sqrt(x+2),y=3-2sqrt(x+2)
Find the zeros of solvefor y,y^2-4x-6y+1=0
The zeros of solvefor y,y^2-4x-6y+1=0 are y=3+2sqrt(x+2),y=3-2sqrt(x+2)