What are the solutions to the equation (x+2)^2=3131-2x ?

The solutions to the equation (x+2)^2=3131-2x are x=53,x=-59

Find the zeros of (x+2)^2=3131-2x

The zeros of (x+2)^2=3131-2x are x=53,x=-59

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(x + 2)^{2} = 3131 - 2 x

x = 53, x = - 59

Solution steps

(x + 2)^{2} = 3131 - 2 x

Expand

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

x^{2} + 4 x + 4 = 3131 - 2 x

Move

2 x

to the left side

x^{2} + 6 x + 4 = 3131

Move

3131

to the left side

x^{2} + 6 x - 3127 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 3127 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 3127) = 112

x_{1, 2} = \frac{- 6 \pm 112}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 112}{2 \cdot 1}, x_{2} = \frac{- 6 - 112}{2 \cdot 1}

x = \frac{- 6 + 112}{2 \cdot 1} : 53

x = \frac{- 6 - 112}{2 \cdot 1} : - 59

The solutions to the quadratic equation are:

x = 53, x = - 59

\frac{5 x - 3}{6} + \frac{x - 5}{18} < \frac{x + 1}{3}

3 w - 15 w - 16 = - 52

3 x \geq 12

- 2 x^{2} + x - 5 = 0

(\frac{- 3 uv}{w ^{5}})^{2}

What are the solutions to the equation (x+2)^2=3131-2x ?
The solutions to the equation (x+2)^2=3131-2x are x=53,x=-59
Find the zeros of (x+2)^2=3131-2x
The zeros of (x+2)^2=3131-2x are x=53,x=-59