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

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

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- 5 - 89}{2 \cdot 4} : \frac{- 5 - 89}{8}

The solutions to the quadratic equation are:

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

3 x^{2} = 39 x

(x + 1)^{2} - 7 (x + 1) + 10 = 0

x^{2} + 35 = 12 x

x, \frac{2 x ^{2}}{3} = d g

2 (x - 1) (x + 3) = (x + 5) (x - 5)

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