What are the solutions to the equation solvefor a,c=ab+a^2 ?

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

Find the zeros of solvefor a,c=ab+a^2

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

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solve for

a, c = ab + a^{2}

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

Solution steps

c = ab + a^{2}

Switch sides

ab + a^{2} = c

Move

c

to the left side

ab + a^{2} - c = 0

Write in the standard form

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

a^{2} + ba - c = 0

Solve with the quadratic formula

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

Simplify

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

a_{1, 2} = \frac{- b \pm b ^{2} + 4 c}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- b + b ^{2} + 4 c}{2 \cdot 1}, a_{2} = \frac{- b - b ^{2} + 4 c}{2 \cdot 1}

a = \frac{- b + b ^{2} + 4 c}{2 \cdot 1} : \frac{- b + b ^{2} + 4 c}{2}

a = \frac{- b - b ^{2} + 4 c}{2 \cdot 1} : \frac{- b - b ^{2} + 4 c}{2}

The solutions to the quadratic equation are:

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

27 x^{2} - \frac{11}{8} x + 5 = 0

(x + 4) (x - 2) = 4 x

x^{2} + 10 x + 23 = - 7

4 x^{2} + 6 x + 1.25 = 0

7 - 8 x - 2 x^{2}

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