What are the solutions to the equation 25x^2+1=20x ?

The solutions to the equation 25x^2+1=20x are x=(2+sqrt(3))/5 ,x=(2-sqrt(3))/5

Find the zeros of 25x^2+1=20x

The zeros of 25x^2+1=20x are x=(2+sqrt(3))/5 ,x=(2-sqrt(3))/5

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25 x^{2} + 1 = 20 x

x = \frac{2 + 3}{5}, x = \frac{2 - 3}{5}

Decimal Notation

x = 0.74641 \dots, x = 0.05358 \dots

Solution steps

25 x^{2} + 1 = 20 x

Move

20 x

to the left side

25 x^{2} + 1 - 20 x = 0

Write in the standard form

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

25 x^{2} - 20 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 25 \cdot 1}{2 \cdot 25}

(- 20)^{2} - 4 \cdot 25 \cdot 1 = 103

x_{1, 2} = \frac{- ( - 20 ) \pm 10 3}{2 \cdot 25}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 10 3}{2 \cdot 25}, x_{2} = \frac{- ( - 20 ) - 10 3}{2 \cdot 25}

x = \frac{- ( - 20 ) + 10 3}{2 \cdot 25} : \frac{2 + 3}{5}

x = \frac{- ( - 20 ) - 10 3}{2 \cdot 25} : \frac{2 - 3}{5}

The solutions to the quadratic equation are:

x = \frac{2 + 3}{5}, x = \frac{2 - 3}{5}

\frac{1}{2} c^{2} + 3 c = 1

t^{2} - t + 6 = 0

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

x, - 493 + 46 x - x^{2} = 0

x^{2} - 9.9 x - 312.5 = 0

What are the solutions to the equation 25x^2+1=20x ?
The solutions to the equation 25x^2+1=20x are x=(2+sqrt(3))/5 ,x=(2-sqrt(3))/5
Find the zeros of 25x^2+1=20x
The zeros of 25x^2+1=20x are x=(2+sqrt(3))/5 ,x=(2-sqrt(3))/5