What are the solutions to the equation 5x=1-3x^2 ?

The solutions to the equation 5x=1-3x^2 are x=-(5+sqrt(37))/6 ,x=(sqrt(37)-5)/6

Find the zeros of 5x=1-3x^2

The zeros of 5x=1-3x^2 are x=-(5+sqrt(37))/6 ,x=(sqrt(37)-5)/6

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5 x = 1 - 3 x^{2}

x = - \frac{5 + 37}{6}, x = \frac{37 - 5}{6}

Decimal Notation

x = - 1.84712 \dots, x = 0.18046 \dots

Solution steps

5 x = 1 - 3 x^{2}

Switch sides

1 - 3 x^{2} = 5 x

Move

5 x

to the left side

1 - 3 x^{2} - 5 x = 0

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) \cdot 1}{2 ( - 3 )}

(- 5)^{2} - 4 (- 3) \cdot 1 = 37

x_{1, 2} = \frac{- ( - 5 ) \pm 37}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 37}{2 ( - 3 )}, x_{2} = \frac{- ( - 5 ) - 37}{2 ( - 3 )}

x = \frac{- ( - 5 ) + 37}{2 ( - 3 )} : - \frac{5 + 37}{6}

x = \frac{- ( - 5 ) - 37}{2 ( - 3 )} : \frac{37 - 5}{6}

The solutions to the quadratic equation are:

x = - \frac{5 + 37}{6}, x = \frac{37 - 5}{6}

2 (\frac{2}{3})^{2} + 5 (\frac{2}{3}) + 6

∣ 2 x + 1 ∣ < 3

0 = - 2 p^{2} + 6000 p

- \frac{π}{4} + 1

x^{2} + 45 x + 20

What are the solutions to the equation 5x=1-3x^2 ?
The solutions to the equation 5x=1-3x^2 are x=-(5+sqrt(37))/6 ,x=(sqrt(37)-5)/6
Find the zeros of 5x=1-3x^2
The zeros of 5x=1-3x^2 are x=-(5+sqrt(37))/6 ,x=(sqrt(37)-5)/6