What are the solutions to the equation solvefor a,y^2+a^2=(a+y)^2-a(a+1) ?

The solutions to the equation solvefor a,y^2+a^2=(a+y)^2-a(a+1) are a=0,a=2y-1

Find the zeros of solvefor a,y^2+a^2=(a+y)^2-a(a+1)

The zeros of solvefor a,y^2+a^2=(a+y)^2-a(a+1) are a=0,a=2y-1

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solve for

a, y^{2} + a^{2} = (a + y)^{2} - a (a + 1)

a = 0, a = 2 y - 1

Solution steps

y^{2} + a^{2} = (a + y)^{2} - a (a + 1)

Expand

(a + y)^{2} - a (a + 1) : 2 a y - a + y^{2}

y^{2} + a^{2} = 2 a y - a + y^{2}

Move

y^{2}

to the left side

a^{2} = 2 a y - a

Subtract

2 a y - a

from both sides

a^{2} - (2 a y - a) = 2 a y - a - (2 a y - a)

Simplify

a^{2} + (- 2 y + 1) a = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 2 y + 1 ) \pm ( - 2 y + 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

Simplify

(- 2 y + 1)^{2} - 4 \cdot 1 \cdot 0 : - 2 y + 1

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

Separate the solutions

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

a = \frac{- ( - 2 y + 1 ) - 2 y + 1}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

a = 0, a = 2 y - 1

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

b, a^{2} + b^{2} = c^{2}

12 x^{2} + 6 x - 6 = 0

\frac{x ^{2}}{5} - \frac{x}{2} = \frac{3}{10}

x^{2} + 24 = 60

What are the solutions to the equation solvefor a,y^2+a^2=(a+y)^2-a(a+1) ?
The solutions to the equation solvefor a,y^2+a^2=(a+y)^2-a(a+1) are a=0,a=2y-1
Find the zeros of solvefor a,y^2+a^2=(a+y)^2-a(a+1)
The zeros of solvefor a,y^2+a^2=(a+y)^2-a(a+1) are a=0,a=2y-1