What are the solutions to the equation solvefor x,x(x-c)=a ?

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

Find the zeros of solvefor x,x(x-c)=a

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

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solve for

x, x (x - c) = a

x = \frac{c + c ^{2} + 4 a}{2}, x = \frac{c - c ^{2} + 4 a}{2}

Solution steps

x (x - c) = a

Expand

x (x - c) : x^{2} - x c

x^{2} - x c = a

Move

a

to the left side

x^{2} - x c - a = 0

Write in the standard form

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

x^{2} - c x - a = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - c ) + c ^{2} + 4 a}{2 \cdot 1}, x_{2} = \frac{- ( - c ) - c ^{2} + 4 a}{2 \cdot 1}

x = \frac{- ( - c ) + c ^{2} + 4 a}{2 \cdot 1} : \frac{c + c ^{2} + 4 a}{2}

x = \frac{- ( - c ) - c ^{2} + 4 a}{2 \cdot 1} : \frac{c - c ^{2} + 4 a}{2}

The solutions to the quadratic equation are:

x = \frac{c + c ^{2} + 4 a}{2}, x = \frac{c - c ^{2} + 4 a}{2}

4 x^{2} + 13 x = 12

2 (x + 3)^{2} + 4 = 0

(x - 11)^{2} = 0

x^{2} - 25 = 25

x^{2} + 4 x + 4 = x - 8

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