What are the solutions to the equation -x^2+11x=28 ?

The solutions to the equation -x^2+11x=28 are x=4,x=7

Find the zeros of -x^2+11x=28

The zeros of -x^2+11x=28 are x=4,x=7

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- x^{2} + 11 x = 28

x = 4, x = 7

Solution steps

- x^{2} + 11 x = 28

Move

28

to the left side

- x^{2} + 11 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 1 ) ( - 28 )}{2 ( - 1 )}

1 1^{2} - 4 (- 1) (- 28) = 3

x_{1, 2} = \frac{- 11 \pm 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 11 + 3}{2 ( - 1 )}, x_{2} = \frac{- 11 - 3}{2 ( - 1 )}

x = \frac{- 11 + 3}{2 ( - 1 )} : 4

x = \frac{- 11 - 3}{2 ( - 1 )} : 7

The solutions to the quadratic equation are:

x = 4, x = 7

18 x - 3 x^{2} = 27

(x - 4)^{3} - (x - 3)^{3} = 343

(x + 2)^{2} = x + 2

(\frac{9}{41})^{2} + y^{2} = 1

1 6^{2} = x^{2} + 6^{2}

What are the solutions to the equation -x^2+11x=28 ?
The solutions to the equation -x^2+11x=28 are x=4,x=7
Find the zeros of -x^2+11x=28
The zeros of -x^2+11x=28 are x=4,x=7