What are the solutions to the equation x(x+4)=12126 ?

The solutions to the equation x(x+4)=12126 are x=-2+sqrt(12130),x=-2-sqrt(12130)

Find the zeros of x(x+4)=12126

The zeros of x(x+4)=12126 are x=-2+sqrt(12130),x=-2-sqrt(12130)

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x (x + 4) = 12126

x = - 2 + 12130, x = - 2 - 12130

Decimal Notation

x = 108.13627 \dots, x = - 112.13627 \dots

Solution steps

x (x + 4) = 12126

Expand

x (x + 4) : x^{2} + 4 x

x^{2} + 4 x = 12126

Move

12126

to the left side

x^{2} + 4 x - 12126 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 12126 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 12126) = 212130

x_{1, 2} = \frac{- 4 \pm 2 12130}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 12130}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 12130}{2 \cdot 1}

x = \frac{- 4 + 2 12130}{2 \cdot 1} : - 2 + 12130

x = \frac{- 4 - 2 12130}{2 \cdot 1} : - 2 - 12130

The solutions to the quadratic equation are:

x = - 2 + 12130, x = - 2 - 12130

x^{2} + x - 12 = 0, x^{1} = 3 x^{2} = - 4

x^{2} - 10 x + 4.8 = 0

(4 x - 3) 2 = x^{2} + (3 x - 3) 2

x (x - 5) - x + 5 = 0

x, v (4 - v (4 + v (4 - x))) = x^{2} - 4

What are the solutions to the equation x(x+4)=12126 ?
The solutions to the equation x(x+4)=12126 are x=-2+sqrt(12130),x=-2-sqrt(12130)
Find the zeros of x(x+4)=12126
The zeros of x(x+4)=12126 are x=-2+sqrt(12130),x=-2-sqrt(12130)