What are the solutions to the equation -x^2-4x+36=0 ?

The solutions to the equation -x^2-4x+36=0 are x=-2(1+sqrt(10)),x=2(sqrt(10)-1)

Find the zeros of -x^2-4x+36=0

The zeros of -x^2-4x+36=0 are x=-2(1+sqrt(10)),x=2(sqrt(10)-1)

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- x^{2} - 4 x + 36 = 0

x = - 2 (1 + 10), x = 2 (10 - 1)

Solution steps

- x^{2} - 4 x + 36 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) \cdot 36 = 410

x_{1, 2} = \frac{- ( - 4 ) \pm 4 10}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 10}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 4 10}{2 ( - 1 )}

x = \frac{- ( - 4 ) + 4 10}{2 ( - 1 )} : - 2 (1 + 10)

x = \frac{- ( - 4 ) - 4 10}{2 ( - 1 )} : 2 (10 - 1)

The solutions to the quadratic equation are:

x = - 2 (1 + 10), x = 2 (10 - 1)

2 x \cdot x^{\frac{1}{2}}

\frac{1}{9} + \frac{1}{8}

(x + 1)^{2} \leq 5 x^{2} + x + 1

(x - 1)^{2} - 13 (x - 1) = - 36

2 x^{2} - 7 x + 5 = 0

What are the solutions to the equation -x^2-4x+36=0 ?
The solutions to the equation -x^2-4x+36=0 are x=-2(1+sqrt(10)),x=2(sqrt(10)-1)
Find the zeros of -x^2-4x+36=0
The zeros of -x^2-4x+36=0 are x=-2(1+sqrt(10)),x=2(sqrt(10)-1)