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

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

Find the zeros of solvefor x,y=x^2-4x

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, y = x^{2} - 4 x

x = 2 + y + 4, x = 2 - y + 4

Solution steps

y = x^{2} - 4 x

Switch sides

x^{2} - 4 x = y

Move

y

to the left side

x^{2} - 4 x - y = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 + y + 4, x = 2 - y + 4

(x - 7)^{2} = 15

j (x) = - 2 x^{2} + 8 x - 9

9 x^{2} + 5 x - 2 = 0

x^{2} - 4 x = 7 x - 28

- 2 r^{2} + 5 r - 22 = - 4 r^{2} + 6 r + 6

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