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

The solutions to the equation 5x^2-4x-4=0 are x=(2(1+sqrt(6)))/5 ,x=(2(1-sqrt(6)))/5

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

The zeros of 5x^2-4x-4=0 are x=(2(1+sqrt(6)))/5 ,x=(2(1-sqrt(6)))/5

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

x = \frac{2 ( 1 + 6 )}{5}, x = \frac{2 ( 1 - 6 )}{5}

Decimal Notation

x = 1.37979 \dots, x = - 0.57979 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 5 (- 4) = 46

x_{1, 2} = \frac{- ( - 4 ) \pm 4 6}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 6}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 4 6}{2 \cdot 5}

x = \frac{- ( - 4 ) + 4 6}{2 \cdot 5} : \frac{2 ( 1 + 6 )}{5}

x = \frac{- ( - 4 ) - 4 6}{2 \cdot 5} : \frac{2 ( 1 - 6 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 1 + 6 )}{5}, x = \frac{2 ( 1 - 6 )}{5}

q, (x + q)^{2} = r

48 + 8 x - x^{2} = 0

r^{2} - 5 r = 0

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

2 x (x + 1) = 0

What are the solutions to the equation 5x^2-4x-4=0 ?
The solutions to the equation 5x^2-4x-4=0 are x=(2(1+sqrt(6)))/5 ,x=(2(1-sqrt(6)))/5
Find the zeros of 5x^2-4x-4=0
The zeros of 5x^2-4x-4=0 are x=(2(1+sqrt(6)))/5 ,x=(2(1-sqrt(6)))/5