What are the solutions to the equation quadraticformula x^2=2x-5 ?

The solutions to the equation quadraticformula x^2=2x-5 are x=1+2i,x=1-2i

Find the zeros of quadraticformula x^2=2x-5

The zeros of quadraticformula x^2=2x-5 are x=1+2i,x=1-2i

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x^{2} = 2 x - 5

x = 1 + 2 i, x = 1 - 2 i

Solution steps

x^{2} = 2 x - 5

Move

5

to the left side

x^{2} + 5 = 2 x

Move

2 x

to the left side

x^{2} + 5 - 2 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 2 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

x_{1, 2} = \frac{- ( - 2 ) \pm 4 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 i}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 4 i}{2 \cdot 1}

x = \frac{- ( - 2 ) + 4 i}{2 \cdot 1} : 1 + 2 i

x = \frac{- ( - 2 ) - 4 i}{2 \cdot 1} : 1 - 2 i

The solutions to the quadratic equation are:

x = 1 + 2 i, x = 1 - 2 i

(- 1)^{1.5}

x, 7 x = 42

n, 8 = n + 8

240

x = \frac{36 \cdot π}{180}

What are the solutions to the equation quadraticformula x^2=2x-5 ?
The solutions to the equation quadraticformula x^2=2x-5 are x=1+2i,x=1-2i
Find the zeros of quadraticformula x^2=2x-5
The zeros of quadraticformula x^2=2x-5 are x=1+2i,x=1-2i