What are the solutions to the equation 2.5x^2+4x+1=0 ?

The solutions to the equation 2.5x^2+4x+1=0 are x=(-4+sqrt(6))/5 ,x=-(4+sqrt(6))/5

Find the zeros of 2.5x^2+4x+1=0

The zeros of 2.5x^2+4x+1=0 are x=(-4+sqrt(6))/5 ,x=-(4+sqrt(6))/5

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2.5 x^{2} + 4 x + 1 = 0

x = \frac{- 4 + 6}{5}, x = - \frac{4 + 6}{5}

Decimal Notation

x = - 0.31010 \dots, x = - 1.28989 \dots

Solution steps

2.5 x^{2} + 4 x + 1 = 0

Multiply both sides by

10

25 x^{2} + 40 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 25 \cdot 10}{2 \cdot 25}

4 0^{2} - 4 \cdot 25 \cdot 10 = 106

x_{1, 2} = \frac{- 40 \pm 10 6}{2 \cdot 25}

Separate the solutions

x_{1} = \frac{- 40 + 10 6}{2 \cdot 25}, x_{2} = \frac{- 40 - 10 6}{2 \cdot 25}

x = \frac{- 40 + 10 6}{2 \cdot 25} : \frac{- 4 + 6}{5}

x = \frac{- 40 - 10 6}{2 \cdot 25} : - \frac{4 + 6}{5}

The solutions to the quadratic equation are:

x = \frac{- 4 + 6}{5}, x = - \frac{4 + 6}{5}

x, y^{2} (a^{2} + x^{2}) = a^{2} x^{2}

(y + k)^{2} - 2 y^{2} - 2 (k + 1) y - 10 = 0

n = 60 h^{2}

x, (x^{2} + 4 x + 20) \cdot y = 23 sin (b) - 15 cos (t)

43 \cdot x^{2} - 2 x - 23 = 0

What are the solutions to the equation 2.5x^2+4x+1=0 ?
The solutions to the equation 2.5x^2+4x+1=0 are x=(-4+sqrt(6))/5 ,x=-(4+sqrt(6))/5
Find the zeros of 2.5x^2+4x+1=0
The zeros of 2.5x^2+4x+1=0 are x=(-4+sqrt(6))/5 ,x=-(4+sqrt(6))/5