What are the solutions to the equation -4x^2+5x+4=0 ?

The solutions to the equation -4x^2+5x+4=0 are x=-(-5+sqrt(89))/8 ,x=(5+sqrt(89))/8

Find the zeros of -4x^2+5x+4=0

The zeros of -4x^2+5x+4=0 are x=-(-5+sqrt(89))/8 ,x=(5+sqrt(89))/8

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

x = - \frac{- 5 + 89}{8}, x = \frac{5 + 89}{8}

Decimal Notation

x = - 0.55424 \dots, x = 1.80424 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 (- 4) \cdot 4 = 89

x_{1, 2} = \frac{- 5 \pm 89}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 5 + 89}{2 ( - 4 )}, x_{2} = \frac{- 5 - 89}{2 ( - 4 )}

x = \frac{- 5 + 89}{2 ( - 4 )} : - \frac{- 5 + 89}{8}

x = \frac{- 5 - 89}{2 ( - 4 )} : \frac{5 + 89}{8}

The solutions to the quadratic equation are:

x = - \frac{- 5 + 89}{8}, x = \frac{5 + 89}{8}

x^{3} + 3 x^{2} - 14 x - 12

16 x + (5 - x) > 30

∣ - 4 x - 3 ∣ \leq 8

- x^{3} + 5 x^{2} - 7 x + 3

(5^{\frac{2}{3}} \cdot 2^{\frac{3}{2}})^{\frac{1}{2}}

What are the solutions to the equation -4x^2+5x+4=0 ?
The solutions to the equation -4x^2+5x+4=0 are x=-(-5+sqrt(89))/8 ,x=(5+sqrt(89))/8
Find the zeros of -4x^2+5x+4=0
The zeros of -4x^2+5x+4=0 are x=-(-5+sqrt(89))/8 ,x=(5+sqrt(89))/8