What is the answer to 1/(2x^2+x)=5 ?

The answer to 1/(2x^2+x)=5 is x=(-5+sqrt(65))/(20),x=(-5-sqrt(65))/(20)

Solve for x: 1/(2x^2+x)=5

The solution is x=(-5+sqrt(65))/(20),x=(-5-sqrt(65))/(20)

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\frac{1}{2 x ^{2} + x} = 5

x = \frac{- 5 + 65}{20}, x = \frac{- 5 - 65}{20}

Decimal Notation

x = 0.15311 \dots, x = - 0.65311 \dots

Solution steps

\frac{1}{2 x ^{2} + x} = 5

Multiply both sides by

2 x^{2} + x

1 = 5 (2 x^{2} + x)

Solve

1 = 5 (2 x^{2} + x) : x = \frac{- 5 + 65}{20}, x = \frac{- 5 - 65}{20}

x = \frac{- 5 + 65}{20}, x = \frac{- 5 - 65}{20}

Verify Solutions

Find undefined (singularity) points:

x = 0, x = - \frac{1}{2}

Combine undefined points with solutions:

x = \frac{- 5 + 65}{20}, x = \frac{- 5 - 65}{20}

\frac{- 5}{4 - x} = 3

\frac{1}{2 x + 5} = 15

\frac{5}{6 x} = \frac{7}{3}

\frac{( x ^{3} - 3 x ^{2} - 13 x + 15 )}{( x - 1 )} = x^{2} - 2 x - 15

a, k = \frac{a ^{2}}{( 1 - a ) ^{2}}

What is the answer to 1/(2x^2+x)=5 ?
The answer to 1/(2x^2+x)=5 is x=(-5+sqrt(65))/(20),x=(-5-sqrt(65))/(20)
Solve for x: 1/(2x^2+x)=5
The solution is x=(-5+sqrt(65))/(20),x=(-5-sqrt(65))/(20)