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

The solutions to the equation solvefor x,y=x^2-5x are x=(5+sqrt(25+4y))/2 ,x=(5-sqrt(25+4y))/2

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

The zeros of solvefor x,y=x^2-5x are x=(5+sqrt(25+4y))/2 ,x=(5-sqrt(25+4y))/2

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solve for

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

x = \frac{5 + 25 + 4 y}{2}, x = \frac{5 - 25 + 4 y}{2}

Solution steps

y = x^{2} - 5 x

Switch sides

x^{2} - 5 x = y

Move

y

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot (- y) : 25 + 4 y

x_{1, 2} = \frac{- ( - 5 ) \pm 25 + 4 y}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 25 + 4 y}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 25 + 4 y}{2 \cdot 1}

x = \frac{- ( - 5 ) + 25 + 4 y}{2 \cdot 1} : \frac{5 + 25 + 4 y}{2}

x = \frac{- ( - 5 ) - 25 + 4 y}{2 \cdot 1} : \frac{5 - 25 + 4 y}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 25 + 4 y}{2}, x = \frac{5 - 25 + 4 y}{2}

30 x - 0.1 x^{2} = 1890, 0 \leq x \leq 150

0 = 7 x^{2} - 31 - 20

Y, Y^{2} - Y = X^{3} - X

x^{2} - 22 x + 105 = 0

x, 6 x^{2} + 18 y^{2} + 36 y + 4 x + 56 = 0

What are the solutions to the equation solvefor x,y=x^2-5x ?
The solutions to the equation solvefor x,y=x^2-5x are x=(5+sqrt(25+4y))/2 ,x=(5-sqrt(25+4y))/2
Find the zeros of solvefor x,y=x^2-5x
The zeros of solvefor x,y=x^2-5x are x=(5+sqrt(25+4y))/2 ,x=(5-sqrt(25+4y))/2