What are the solutions to the equation x^2=4x-9 ?

The solutions to the equation x^2=4x-9 are x=2+sqrt(5)i,x=2-sqrt(5)i

Find the zeros of x^2=4x-9

The zeros of x^2=4x-9 are x=2+sqrt(5)i,x=2-sqrt(5)i

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x^{2} = 4 x - 9

x = 2 + 5 i, x = 2 - 5 i

Solution steps

x^{2} = 4 x - 9

Move

9

to the left side

x^{2} + 9 = 4 x

Move

4 x

to the left side

x^{2} + 9 - 4 x = 0

Write in the standard form

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

x^{2} - 4 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 9 : 25 i

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 5 i}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 5 i}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 5 i}{2 \cdot 1} : 2 + 5 i

x = \frac{- ( - 4 ) - 2 5 i}{2 \cdot 1} : 2 - 5 i

The solutions to the quadratic equation are:

x = 2 + 5 i, x = 2 - 5 i

10 x^{5} y + 2 x^{3} y^{2} + 14 x^{2} y^{4}

i^{- 3}

\frac{12}{30} \cdot \frac{2}{30} \cdot \frac{17}{29}

5 x - 3 < 42 and - 5 x + 8 < 58

8 - 6 x^{2} = 0

What are the solutions to the equation x^2=4x-9 ?
The solutions to the equation x^2=4x-9 are x=2+sqrt(5)i,x=2-sqrt(5)i
Find the zeros of x^2=4x-9
The zeros of x^2=4x-9 are x=2+sqrt(5)i,x=2-sqrt(5)i