What are the solutions to the equation 5=-x^2+10x-21 ?

The solutions to the equation 5=-x^2+10x-21 are x=5-i,x=5+i

Find the zeros of 5=-x^2+10x-21

The zeros of 5=-x^2+10x-21 are x=5-i,x=5+i

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5 = - x^{2} + 10 x - 21

x = 5 - i, x = 5 + i

Solution steps

5 = - x^{2} + 10 x - 21

Switch sides

- x^{2} + 10 x - 21 = 5

Move

5

to the left side

- x^{2} + 10 x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 1 ) ( - 26 )}{2 ( - 1 )}

Simplify

1 0^{2} - 4 (- 1) (- 26) : 2 i

x_{1, 2} = \frac{- 10 \pm 2 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 10 + 2 i}{2 ( - 1 )}, x_{2} = \frac{- 10 - 2 i}{2 ( - 1 )}

x = \frac{- 10 + 2 i}{2 ( - 1 )} : 5 - i

x = \frac{- 10 - 2 i}{2 ( - 1 )} : 5 + i

The solutions to the quadratic equation are:

x = 5 - i, x = 5 + i

r^{2} + 2 r - 8 = 0

- 5 x < 25

2 0^{- 8} \cdot 2 0^{8}

x^{2} - 21 = 4 x

3 x^{- 4} y^{9} \cdot 8 x^{- 1} y^{- 6}

What are the solutions to the equation 5=-x^2+10x-21 ?
The solutions to the equation 5=-x^2+10x-21 are x=5-i,x=5+i
Find the zeros of 5=-x^2+10x-21
The zeros of 5=-x^2+10x-21 are x=5-i,x=5+i