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

The solutions to the equation 5x^2-5=x are x=(1+sqrt(101))/(10),x=(1-sqrt(101))/(10)

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

The zeros of 5x^2-5=x are x=(1+sqrt(101))/(10),x=(1-sqrt(101))/(10)

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5 x^{2} - 5 = x

x = \frac{1 + 101}{10}, x = \frac{1 - 101}{10}

Decimal Notation

x = 1.10498 \dots, x = - 0.90498 \dots

Solution steps

5 x^{2} - 5 = x

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 5 (- 5) = 101

x_{1, 2} = \frac{- ( - 1 ) \pm 101}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 101}{2 \cdot 5}, x_{2} = \frac{- ( - 1 ) - 101}{2 \cdot 5}

x = \frac{- ( - 1 ) + 101}{2 \cdot 5} : \frac{1 + 101}{10}

x = \frac{- ( - 1 ) - 101}{2 \cdot 5} : \frac{1 - 101}{10}

The solutions to the quadratic equation are:

x = \frac{1 + 101}{10}, x = \frac{1 - 101}{10}

m^{2} + 4 m = 32

5 x^{2} - 4 = x

y, s (y) = 3 yy + 5

2 y^{2} + 2 y - 4 = 0

20 x^{2} + 4 x = 0

What are the solutions to the equation 5x^2-5=x ?
The solutions to the equation 5x^2-5=x are x=(1+sqrt(101))/(10),x=(1-sqrt(101))/(10)
Find the zeros of 5x^2-5=x
The zeros of 5x^2-5=x are x=(1+sqrt(101))/(10),x=(1-sqrt(101))/(10)