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

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

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

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

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

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

Decimal Notation

x = - 0.42539 \dots, x = 1.17539 \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{3 + 41}{8}

The solutions to the quadratic equation are:

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

x^{2} + 7 x - 30 = 0

x, s (x) = (x^{2} - 2 x - 1) - (- 2 x^{2} - 7 x + 12)

5 w^{2} - 90 = 0

x, x^{2} + 5 = 0

0.32 x^{2} + 37 x - 627 = 0

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