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

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

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

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

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- x^{2} + 13 x = 36

x = 4, x = 9

Solution steps

- x^{2} + 13 x = 36

Move

36

to the left side

- 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

x - 13 = 27

- 2 x^{2} + 16 x - 24 = 0

\frac{7}{8} - \frac{3}{8}

\frac{3}{\frac{2}{5}}

4 x^{2} + 2 x + 4 = 0

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