What are the solutions to the equation -x^2-3x+18=0 ?

The solutions to the equation -x^2-3x+18=0 are x=-6,x=3

Find the zeros of -x^2-3x+18=0

The zeros of -x^2-3x+18=0 are x=-6,x=3

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- x^{2} - 3 x + 18 = 0

x = - 6, x = 3

Solution steps

- x^{2} - 3 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) \cdot 18}{2 ( - 1 )}

(- 3)^{2} - 4 (- 1) \cdot 18 = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 ( - 1 )}, x_{2} = \frac{- ( - 3 ) - 9}{2 ( - 1 )}

x = \frac{- ( - 3 ) + 9}{2 ( - 1 )} : - 6

x = \frac{- ( - 3 ) - 9}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

x = - 6, x = 3

4 t^{2} - 9 t + 5 = 0

3 k^{2} - 18 k - 21 = 0

m^{2} - 6 m - 55 = 0

0 = 6 x^{2} - 6 x - 72

25 x^{2} - 6 x = 0

What are the solutions to the equation -x^2-3x+18=0 ?
The solutions to the equation -x^2-3x+18=0 are x=-6,x=3
Find the zeros of -x^2-3x+18=0
The zeros of -x^2-3x+18=0 are x=-6,x=3