What are the solutions to the equation x^2+20x-4=0 ?

The solutions to the equation x^2+20x-4=0 are x=2(sqrt(26)-5),x=-2(5+sqrt(26))

Find the zeros of x^2+20x-4=0

The zeros of x^2+20x-4=0 are x=2(sqrt(26)-5),x=-2(5+sqrt(26))

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x^{2} + 20 x - 4 = 0

x = 2 (26 - 5), x = - 2 (5 + 26)

Decimal

x = 0.19803 \dots, x = - 20.19803 \dots

Solution steps

x^{2} + 20 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 4) = 426

x_{1, 2} = \frac{- 20 \pm 4 26}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 4 26}{2 \cdot 1}, x_{2} = \frac{- 20 - 4 26}{2 \cdot 1}

x = \frac{- 20 + 4 26}{2 \cdot 1} : 2 (26 - 5)

x = \frac{- 20 - 4 26}{2 \cdot 1} : - 2 (5 + 26)

The solutions to the quadratic equation are:

x = 2 (26 - 5), x = - 2 (5 + 26)

3 (3 x + 5)^{2} - 36 = - 21

x^{2} + 4 x = 8 x + 5

2 x^{2} - 1 \frac{7}{25} = 0

a x^{2} + 2 x + 1 = 0

2 x^{2} + 3 x - 1

What are the solutions to the equation x^2+20x-4=0 ?
The solutions to the equation x^2+20x-4=0 are x=2(sqrt(26)-5),x=-2(5+sqrt(26))
Find the zeros of x^2+20x-4=0
The zeros of x^2+20x-4=0 are x=2(sqrt(26)-5),x=-2(5+sqrt(26))