What are the solutions to the equation solvefor y=x^2-5x+6,x ?

The solutions to the equation solvefor y=x^2-5x+6,x are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2

Find the zeros of solvefor y=x^2-5x+6,x

The zeros of solvefor y=x^2-5x+6,x are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

y = x^{2} - 5 x + 6, x

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

Solution steps

y = x^{2} - 5 x + 6

Switch sides

x^{2} - 5 x + 6 = y

Move

y

to the left side

x^{2} - 5 x + 6 - y = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( 6 - y )}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot (6 - y) : 4 y + 1

x_{1, 2} = \frac{- ( - 5 ) \pm 4 y + 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 4 y + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 4 y + 1}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

x, f (x + 5) = x^{2} + 2 x + 3

(2 x - 1) (2 x + 1) = (2 x - 3) 2

x, f^{1} (x) = x^{2} + 2

2 y (y + 5) - 5 = 0

y^{2} + 18 y - 40 = 0

What are the solutions to the equation solvefor y=x^2-5x+6,x ?
The solutions to the equation solvefor y=x^2-5x+6,x are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2
Find the zeros of solvefor y=x^2-5x+6,x
The zeros of solvefor y=x^2-5x+6,x are x=(5+sqrt(4y+1))/2 ,x=(5-sqrt(4y+1))/2