What are the solutions to the equation A=x^2+x-2 ?

The solutions to the equation A=x^2+x-2 are x=(-1+sqrt(4A+9))/2 ,x=(-1-sqrt(4A+9))/2

Find the zeros of A=x^2+x-2

The zeros of A=x^2+x-2 are x=(-1+sqrt(4A+9))/2 ,x=(-1-sqrt(4A+9))/2

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A = x^{2} + x - 2

x = \frac{- 1 + 4 A + 9}{2}, x = \frac{- 1 - 4 A + 9}{2}

Solution steps

A = x^{2} + x - 2

Switch sides

x^{2} + x - 2 = A

Move

A

to the left side

x^{2} + x - 2 - A = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot (- 2 - A) : 4 A + 9

x_{1, 2} = \frac{- 1 \pm 4 A + 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 4 A + 9}{2 \cdot 1}, x_{2} = \frac{- 1 - 4 A + 9}{2 \cdot 1}

x = \frac{- 1 + 4 A + 9}{2 \cdot 1} : \frac{- 1 + 4 A + 9}{2}

x = \frac{- 1 - 4 A + 9}{2 \cdot 1} : \frac{- 1 - 4 A + 9}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 4 A + 9}{2}, x = \frac{- 1 - 4 A + 9}{2}

8 x^{2} - 28 x - 60 = 0

r^{2} + \frac{4}{3} r = \frac{4}{3}

c, y^{2} + 2 x^{2} = c^{2}

2 = (1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1

- 12 x = 12 x^{2} + 3

What are the solutions to the equation A=x^2+x-2 ?
The solutions to the equation A=x^2+x-2 are x=(-1+sqrt(4A+9))/2 ,x=(-1-sqrt(4A+9))/2
Find the zeros of A=x^2+x-2
The zeros of A=x^2+x-2 are x=(-1+sqrt(4A+9))/2 ,x=(-1-sqrt(4A+9))/2