What are the solutions to the equation x^2+1=5x ?

The solutions to the equation x^2+1=5x are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2

Find the zeros of x^2+1=5x

The zeros of x^2+1=5x are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2

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x^{2} + 1 = 5 x

x = \frac{5 + 21}{2}, x = \frac{5 - 21}{2}

Decimal Notation

x = 4.79128 \dots, x = 0.20871 \dots

Solution steps

x^{2} + 1 = 5 x

Move

5 x

to the left side

x^{2} + 1 - 5 x = 0

Write in the standard form

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

x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 1 \cdot 1 = 21

x_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 1}

x = \frac{- ( - 5 ) + 21}{2 \cdot 1} : \frac{5 + 21}{2}

x = \frac{- ( - 5 ) - 21}{2 \cdot 1} : \frac{5 - 21}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 21}{2}, x = \frac{5 - 21}{2}

\frac{2}{3} + \frac{3 k}{4} = \frac{71}{12}

- 2 x (y - 1) + 3 (1 - y)

3 (4 - 5)^{2} + 4 (5 - 1) - 3 (1 + 2)^{2} (2) + 2 (3) (3)

- 5 \leq 4 - 9 x < - 2

4 m^{2} - 37 m + 9

What are the solutions to the equation x^2+1=5x ?
The solutions to the equation x^2+1=5x are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2
Find the zeros of x^2+1=5x
The zeros of x^2+1=5x are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2