What are the solutions to the equation -4x+5=(x-2)^2+1 ?

The solutions to the equation -4x+5=(x-2)^2+1 are x=0

Find the zeros of -4x+5=(x-2)^2+1

The zeros of -4x+5=(x-2)^2+1 are x=0

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- 4 x + 5 = (x - 2)^{2} + 1

x = 0

Solution steps

- 4 x + 5 = (x - 2)^{2} + 1

Expand

(x - 2)^{2} + 1 : x^{2} - 4 x + 5

- 4 x + 5 = x^{2} - 4 x + 5

Switch sides

x^{2} - 4 x + 5 = - 4 x + 5

Move

5

to the right side

x^{2} - 4 x = - 4 x

Move

4 x

to the left side

x^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

x = 0

(\frac{- 1}{4})^{- (2 + 3)}

- 9 - 3 (7 - 2) + \frac{24}{3}

\frac{35 y ^{8} x ^{5}}{- 21 y ^{- 1} x ^{9}}

∣ 3 ∣ \leq ∣ - 1 ∣

What are the solutions to the equation -4x+5=(x-2)^2+1 ?
The solutions to the equation -4x+5=(x-2)^2+1 are x=0
Find the zeros of -4x+5=(x-2)^2+1
The zeros of -4x+5=(x-2)^2+1 are x=0