What are the solutions to the equation -x^2-x+42=0 ?

The solutions to the equation -x^2-x+42=0 are x=-7,x=6

Find the zeros of -x^2-x+42=0

The zeros of -x^2-x+42=0 are x=-7,x=6

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- x^{2} - x + 42 = 0

x = - 7, x = 6

Solution steps

- x^{2} - x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 42}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 42 = 13

x_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 13}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 13}{2 ( - 1 )}

x = \frac{- ( - 1 ) + 13}{2 ( - 1 )} : - 7

x = \frac{- ( - 1 ) - 13}{2 ( - 1 )} : 6

The solutions to the quadratic equation are:

x = - 7, x = 6

(x - 3)^{2} - 2 (x - 3) + 1 = 0

- 0.001 x^{2} - 0.4 x + 210 = 0.0006 x^{2} + 0.26 x + 128

(x + 3)^{2} - 9 = 7

x^{2} + 6 x - 16 = 0

(1 - x)^{2} - 9 = 0

What are the solutions to the equation -x^2-x+42=0 ?
The solutions to the equation -x^2-x+42=0 are x=-7,x=6
Find the zeros of -x^2-x+42=0
The zeros of -x^2-x+42=0 are x=-7,x=6