What are the solutions to the equation -x^2-2x+24=0 ?

The solutions to the equation -x^2-2x+24=0 are x=-6,x=4

Find the zeros of -x^2-2x+24=0

The zeros of -x^2-2x+24=0 are x=-6,x=4

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- x^{2} - 2 x + 24 = 0

x = - 6, x = 4

Solution steps

- x^{2} - 2 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 24}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) \cdot 24 = 10

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 10}{2 ( - 1 )} : - 6

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

The solutions to the quadratic equation are:

x = - 6, x = 4

x^{2} - 4 + 6 = 0

x^{2} = - 5 x - 3

x^{2} = lo g_{2} (48) - lo g_{2} (3)

x^{2} = 4 x - 13

16 r^{2} + 18 r + 5 = 0

What are the solutions to the equation -x^2-2x+24=0 ?
The solutions to the equation -x^2-2x+24=0 are x=-6,x=4
Find the zeros of -x^2-2x+24=0
The zeros of -x^2-2x+24=0 are x=-6,x=4