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

The solutions to the equation 2x^2+1=5x are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

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

The zeros of 2x^2+1=5x are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

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2 x^{2} + 1 = 5 x

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

Decimal Notation

x = 2.28077 \dots, x = 0.21922 \dots

Solution steps

2 x^{2} + 1 = 5 x

Move

5 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

x_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

x = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

x = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

∣ 3 - 2 x ∣ = 3

2 x + 8 < 25

∣ 2 x + 1 ∣ = 17

(7 v - 2) \frac{1}{4} + 12 = 7

\frac{x + 4}{x + 3} < \frac{x - 1}{x - 4}

What are the solutions to the equation 2x^2+1=5x ?
The solutions to the equation 2x^2+1=5x are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4
Find the zeros of 2x^2+1=5x
The zeros of 2x^2+1=5x are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4