What are the solutions to the equation 2x=5-4x^2 ?

The solutions to the equation 2x=5-4x^2 are x=-(1+sqrt(21))/4 ,x=(sqrt(21)-1)/4

Find the zeros of 2x=5-4x^2

The zeros of 2x=5-4x^2 are x=-(1+sqrt(21))/4 ,x=(sqrt(21)-1)/4

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2 x = 5 - 4 x^{2}

x = - \frac{1 + 21}{4}, x = \frac{21 - 1}{4}

Decimal Notation

x = - 1.39564 \dots, x = 0.89564 \dots

Solution steps

2 x = 5 - 4 x^{2}

Switch sides

5 - 4 x^{2} = 2 x

Move

2 x

to the left side

5 - 4 x^{2} - 2 x = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 (- 4) \cdot 5 = 221

x_{1, 2} = \frac{- ( - 2 ) \pm 2 21}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 21}{2 ( - 4 )}, x_{2} = \frac{- ( - 2 ) - 2 21}{2 ( - 4 )}

x = \frac{- ( - 2 ) + 2 21}{2 ( - 4 )} : - \frac{1 + 21}{4}

x = \frac{- ( - 2 ) - 2 21}{2 ( - 4 )} : \frac{21 - 1}{4}

The solutions to the quadratic equation are:

x = - \frac{1 + 21}{4}, x = \frac{21 - 1}{4}

x - \frac{5}{2} < \frac{x}{4} - 6

2 - \frac{b}{3} = - \frac{5}{2}

\frac{( \frac{1}{y} + \frac{1}{x} )}{\frac{x}{y} - \frac{y}{x}}

- n + (- 4) - (- 4 n) + 6

2 (3 x + 3) - 4 (5 x - 3) = x (x - 3) - x (x + 5)

What are the solutions to the equation 2x=5-4x^2 ?
The solutions to the equation 2x=5-4x^2 are x=-(1+sqrt(21))/4 ,x=(sqrt(21)-1)/4
Find the zeros of 2x=5-4x^2
The zeros of 2x=5-4x^2 are x=-(1+sqrt(21))/4 ,x=(sqrt(21)-1)/4