What are the solutions to the equation -x^2+13x-42=0 ?

The solutions to the equation -x^2+13x-42=0 are x=6,x=7

Find the zeros of -x^2+13x-42=0

The zeros of -x^2+13x-42=0 are x=6,x=7

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- x^{2} + 13 x - 42 = 0

x = 6, x = 7

Solution steps

- x^{2} + 13 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 1 ) ( - 42 )}{2 ( - 1 )}

1 3^{2} - 4 (- 1) (- 42) = 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 6, x = 7

x^{2} - 3 x - \frac{7}{4} = 0

(2 x + 1) 3 x + 6 x (2 x + 1) = - 9 x (1 + 2 x)

x, (C - x) (1 + k x) = B (k q x)

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

0 = - 2 x^{2} - 16 x - 30

What are the solutions to the equation -x^2+13x-42=0 ?
The solutions to the equation -x^2+13x-42=0 are x=6,x=7
Find the zeros of -x^2+13x-42=0
The zeros of -x^2+13x-42=0 are x=6,x=7