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

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

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

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

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

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

Solution steps

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

Switch sides

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)

7 + x \leq 2 (x + 3) + 2

\frac{4 x - 3}{2} - \frac{3 x - 3}{3} < 3

3 x^{2} + 2 x - 26 = 0

- 41 a - 4 = 47 a - 3

\frac{3 ^{6}}{3 ^{2}}

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