What are the solutions to the equation -x^2+13x-36=0 ?

The solutions to the equation -x^2+13x-36=0 are x=4,x=9

Find the zeros of -x^2+13x-36=0

The zeros of -x^2+13x-36=0 are x=4,x=9

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- x^{2} + 13 x - 36 = 0

x = 4, x = 9

Solution steps

- x^{2} + 13 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 1 ) ( - 36 )}{2 ( - 1 )}

1 3^{2} - 4 (- 1) (- 36) = 5

x_{1, 2} = \frac{- 13 \pm 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 13 + 5}{2 ( - 1 )}, x_{2} = \frac{- 13 - 5}{2 ( - 1 )}

x = \frac{- 13 + 5}{2 ( - 1 )} : 4

x = \frac{- 13 - 5}{2 ( - 1 )} : 9

The solutions to the quadratic equation are:

x = 4, x = 9

lo g_{4} (8)

2 - 16 t = 6 (- 3 t + 2)

12 > - 2 x + 6 > - 2

3 y + 4 (x + 2 y)

36 - 46

What are the solutions to the equation -x^2+13x-36=0 ?
The solutions to the equation -x^2+13x-36=0 are x=4,x=9
Find the zeros of -x^2+13x-36=0
The zeros of -x^2+13x-36=0 are x=4,x=9