What are the solutions to the equation x^2+4x-136=0 ?

The solutions to the equation x^2+4x-136=0 are x=2(sqrt(35)-1),x=-2(1+sqrt(35))

Find the zeros of x^2+4x-136=0

The zeros of x^2+4x-136=0 are x=2(sqrt(35)-1),x=-2(1+sqrt(35))

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x^{2} + 4 x - 136 = 0

x = 2 (35 - 1), x = - 2 (1 + 35)

Solution steps

x^{2} + 4 x - 136 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 136 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 136) = 435

x_{1, 2} = \frac{- 4 \pm 4 35}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 4 35}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 35}{2 \cdot 1}

x = \frac{- 4 + 4 35}{2 \cdot 1} : 2 (35 - 1)

x = \frac{- 4 - 4 35}{2 \cdot 1} : - 2 (1 + 35)

The solutions to the quadratic equation are:

x = 2 (35 - 1), x = - 2 (1 + 35)

2 x^{2} + 3 x y - 2 y^{2}

4 x - 7 = 21

x + 1 > - 2

\frac{10 !}{5 ! ( 5 ! )}

6 x^{2} - 4 x - 2

What are the solutions to the equation x^2+4x-136=0 ?
The solutions to the equation x^2+4x-136=0 are x=2(sqrt(35)-1),x=-2(1+sqrt(35))
Find the zeros of x^2+4x-136=0
The zeros of x^2+4x-136=0 are x=2(sqrt(35)-1),x=-2(1+sqrt(35))