What are the solutions to the equation x^2+22x=4 ?

The solutions to the equation x^2+22x=4 are x=-11+5sqrt(5),x=-11-5sqrt(5)

Find the zeros of x^2+22x=4

The zeros of x^2+22x=4 are x=-11+5sqrt(5),x=-11-5sqrt(5)

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x^{2} + 22 x = 4

x = - 11 + 55, x = - 11 - 55

Solution steps

x^{2} + 22 x = 4

Move

4

to the left side

x^{2} + 22 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

2 2^{2} - 4 \cdot 1 \cdot (- 4) = 105

x_{1, 2} = \frac{- 22 \pm 10 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 22 + 10 5}{2 \cdot 1}, x_{2} = \frac{- 22 - 10 5}{2 \cdot 1}

x = \frac{- 22 + 10 5}{2 \cdot 1} : - 11 + 55

x = \frac{- 22 - 10 5}{2 \cdot 1} : - 11 - 55

The solutions to the quadratic equation are:

x = - 11 + 55, x = - 11 - 55

3 a = a^{2}

4 c + 7 = 2 c - 1

(- 5 b^{3}) (- 3 b^{6})

- 16 t + 8 t + 80

∣ - 2 x + 3 ∣ \leq 15

What are the solutions to the equation x^2+22x=4 ?
The solutions to the equation x^2+22x=4 are x=-11+5sqrt(5),x=-11-5sqrt(5)
Find the zeros of x^2+22x=4
The zeros of x^2+22x=4 are x=-11+5sqrt(5),x=-11-5sqrt(5)