What are the solutions to the equation 2-3x-4x^2=0 ?

The solutions to the equation 2-3x-4x^2=0 are x=-(3+sqrt(41))/8 ,x=(sqrt(41)-3)/8

Find the zeros of 2-3x-4x^2=0

The zeros of 2-3x-4x^2=0 are x=-(3+sqrt(41))/8 ,x=(sqrt(41)-3)/8

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2 - 3 x - 4 x^{2} = 0

x = - \frac{3 + 41}{8}, x = \frac{41 - 3}{8}

Decimal Notation

x = - 1.17539 \dots, x = 0.42539 \dots

Solution steps

2 - 3 x - 4 x^{2} = 0

Write in the standard form

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

- 4 x^{2} - 3 x + 2 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 4) \cdot 2 = 41

x_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 41}{2 ( - 4 )}, x_{2} = \frac{- ( - 3 ) - 41}{2 ( - 4 )}

x = \frac{- ( - 3 ) + 41}{2 ( - 4 )} : - \frac{3 + 41}{8}

x = \frac{- ( - 3 ) - 41}{2 ( - 4 )} : \frac{41 - 3}{8}

The solutions to the quadratic equation are:

x = - \frac{3 + 41}{8}, x = \frac{41 - 3}{8}

x^{2} + \frac{3}{3} x + \frac{7}{3} = 0

- 2 x^{2} - 17 x - 53 = 0

x, 3 y^{2} - 12 z^{2} - 4 x^{2} = 12

25 x^{2} - 20 x + 9 = 0

4 x^{2} + 6 x - 86 = 2

What are the solutions to the equation 2-3x-4x^2=0 ?
The solutions to the equation 2-3x-4x^2=0 are x=-(3+sqrt(41))/8 ,x=(sqrt(41)-3)/8
Find the zeros of 2-3x-4x^2=0
The zeros of 2-3x-4x^2=0 are x=-(3+sqrt(41))/8 ,x=(sqrt(41)-3)/8