What are the solutions to the equation 5x=1-2x^2 ?

The solutions to the equation 5x=1-2x^2 are x=-(5+sqrt(33))/4 ,x=(sqrt(33)-5)/4

Find the zeros of 5x=1-2x^2

The zeros of 5x=1-2x^2 are x=-(5+sqrt(33))/4 ,x=(sqrt(33)-5)/4

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5 x = 1 - 2 x^{2}

x = - \frac{5 + 33}{4}, x = \frac{33 - 5}{4}

Decimal Notation

x = - 2.68614 \dots, x = 0.18614 \dots

Solution steps

5 x = 1 - 2 x^{2}

Switch sides

1 - 2 x^{2} = 5 x

Move

5 x

to the left side

1 - 2 x^{2} - 5 x = 0

Write in the standard form

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

- 2 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 (- 2) \cdot 1 = 33

x_{1, 2} = \frac{- ( - 5 ) \pm 33}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 33}{2 ( - 2 )}, x_{2} = \frac{- ( - 5 ) - 33}{2 ( - 2 )}

x = \frac{- ( - 5 ) + 33}{2 ( - 2 )} : - \frac{5 + 33}{4}

x = \frac{- ( - 5 ) - 33}{2 ( - 2 )} : \frac{33 - 5}{4}

The solutions to the quadratic equation are:

x = - \frac{5 + 33}{4}, x = \frac{33 - 5}{4}

x^{2} + x = \frac{4}{3}

m^{2} + 50 m - 3 = 0

x, f (x + 3) = 3 x^{2} - x + 7

10 x^{2} + 11 x - 3 = 0

x^{2} - \frac{1}{8} x + 23 = 100

What are the solutions to the equation 5x=1-2x^2 ?
The solutions to the equation 5x=1-2x^2 are x=-(5+sqrt(33))/4 ,x=(sqrt(33)-5)/4
Find the zeros of 5x=1-2x^2
The zeros of 5x=1-2x^2 are x=-(5+sqrt(33))/4 ,x=(sqrt(33)-5)/4