What are the solutions to the equation x^2+8x-36=0 ?

The solutions to the equation x^2+8x-36=0 are x=2(sqrt(13)-2),x=-2(2+sqrt(13))

Find the zeros of x^2+8x-36=0

The zeros of x^2+8x-36=0 are x=2(sqrt(13)-2),x=-2(2+sqrt(13))

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x^{2} + 8 x - 36 = 0

x = 2 (13 - 2), x = - 2 (2 + 13)

Solution steps

x^{2} + 8 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 36) = 413

x_{1, 2} = \frac{- 8 \pm 4 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4 13}{2 \cdot 1}, x_{2} = \frac{- 8 - 4 13}{2 \cdot 1}

x = \frac{- 8 + 4 13}{2 \cdot 1} : 2 (13 - 2)

x = \frac{- 8 - 4 13}{2 \cdot 1} : - 2 (2 + 13)

The solutions to the quadratic equation are:

x = 2 (13 - 2), x = - 2 (2 + 13)

20 + 17 v = - 6 + 15 v

2 (x + 1) + 2 x = 4 x + 2

25 p^{2} + 30 p + 9

6 x = - 18

5 > 2 - 9 x > - 4

What are the solutions to the equation x^2+8x-36=0 ?
The solutions to the equation x^2+8x-36=0 are x=2(sqrt(13)-2),x=-2(2+sqrt(13))
Find the zeros of x^2+8x-36=0
The zeros of x^2+8x-36=0 are x=2(sqrt(13)-2),x=-2(2+sqrt(13))