What are the solutions to the equation -x^2+20x-126=5 ?

The solutions to the equation -x^2+20x-126=5 are x=10-sqrt(31)i,x=10+sqrt(31)i

Find the zeros of -x^2+20x-126=5

The zeros of -x^2+20x-126=5 are x=10-sqrt(31)i,x=10+sqrt(31)i

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- x^{2} + 20 x - 126 = 5

x = 10 - 31 i, x = 10 + 31 i

Solution steps

- x^{2} + 20 x - 126 = 5

Move

5

to the left side

- x^{2} + 20 x - 131 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 1 ) ( - 131 )}{2 ( - 1 )}

Simplify

2 0^{2} - 4 (- 1) (- 131) : 231 i

x_{1, 2} = \frac{- 20 \pm 2 31 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 20 + 2 31 i}{2 ( - 1 )}, x_{2} = \frac{- 20 - 2 31 i}{2 ( - 1 )}

x = \frac{- 20 + 2 31 i}{2 ( - 1 )} : 10 - 31 i

x = \frac{- 20 - 2 31 i}{2 ( - 1 )} : 10 + 31 i

The solutions to the quadratic equation are:

x = 10 - 31 i, x = 10 + 31 i

15 x^{2} = 135

- 2 = x^{2}

(x - 21)^{2} = 25

20 n^{2} - 9 n - 20 = 0

y^{2} - 121 = 0

What are the solutions to the equation -x^2+20x-126=5 ?
The solutions to the equation -x^2+20x-126=5 are x=10-sqrt(31)i,x=10+sqrt(31)i
Find the zeros of -x^2+20x-126=5
The zeros of -x^2+20x-126=5 are x=10-sqrt(31)i,x=10+sqrt(31)i