What are the solutions to the equation x^2+4x-1=7 ?

The solutions to the equation x^2+4x-1=7 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))

Find the zeros of x^2+4x-1=7

The zeros of x^2+4x-1=7 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))

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x^{2} + 4 x - 1 = 7

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

Decimal Notation

x = 1.46410 \dots, x = - 5.46410 \dots

Solution steps

x^{2} + 4 x - 1 = 7

Move

7

to the left side

x^{2} + 4 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 8) = 43

x_{1, 2} = \frac{- 4 \pm 4 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 4 3}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 3}{2 \cdot 1}

x = \frac{- 4 + 4 3}{2 \cdot 1} : 2 (3 - 1)

x = \frac{- 4 - 4 3}{2 \cdot 1} : - 2 (1 + 3)

The solutions to the quadratic equation are:

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

c^{1} + c^{2} = - 2

x, x^{2} + k x + 36 = 0

(x + 1)^{2} - 8 \cdot (x + 1) + 7 = 0

x, x^{2} + (y - 32 x)^{2} = 1

x^{2} - x - 48 = 0

What are the solutions to the equation x^2+4x-1=7 ?
The solutions to the equation x^2+4x-1=7 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))
Find the zeros of x^2+4x-1=7
The zeros of x^2+4x-1=7 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))