What are the solutions to the equation 4x^2+8x+4.1=0 ?

The solutions to the equation 4x^2+8x+4.1=0 are x=-1+i(sqrt(10))/(20),x=-1-i(sqrt(10))/(20)

Find the zeros of 4x^2+8x+4.1=0

The zeros of 4x^2+8x+4.1=0 are x=-1+i(sqrt(10))/(20),x=-1-i(sqrt(10))/(20)

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4 x^{2} + 8 x + 4.1 = 0

x = - 1 + i \frac{10}{20}, x = - 1 - i \frac{10}{20}

Solution steps

4 x^{2} + 8 x + 4.1 = 0

Multiply both sides by

10

40 x^{2} + 80 x + 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 40 \cdot 41}{2 \cdot 40}

Simplify

8 0^{2} - 4 \cdot 40 \cdot 41 : 410 i

x_{1, 2} = \frac{- 80 \pm 4 10 i}{2 \cdot 40}

Separate the solutions

x_{1} = \frac{- 80 + 4 10 i}{2 \cdot 40}, x_{2} = \frac{- 80 - 4 10 i}{2 \cdot 40}

x = \frac{- 80 + 4 10 i}{2 \cdot 40} : - 1 + i \frac{10}{20}

x = \frac{- 80 - 4 10 i}{2 \cdot 40} : - 1 - i \frac{10}{20}

The solutions to the quadratic equation are:

x = - 1 + i \frac{10}{20}, x = - 1 - i \frac{10}{20}

∣ x - 4 ∣ \geq 7

\frac{( 1 + 0.1 ) ^{5} - 1}{0.1 ( 0.1 + 1 ) ^{5}}

18 - 5 (2 x + 3) = 9 - 15 x

\frac{5 !}{1}

\frac{6 ^{5}}{6 ^{4}}

What are the solutions to the equation 4x^2+8x+4.1=0 ?
The solutions to the equation 4x^2+8x+4.1=0 are x=-1+i(sqrt(10))/(20),x=-1-i(sqrt(10))/(20)
Find the zeros of 4x^2+8x+4.1=0
The zeros of 4x^2+8x+4.1=0 are x=-1+i(sqrt(10))/(20),x=-1-i(sqrt(10))/(20)