What are the solutions to the equation completesquare x^2+4x+7=5 ?

The solutions to the equation completesquare x^2+4x+7=5 are x=sqrt(2)-2,x=-sqrt(2)-2

Find the zeros of completesquare x^2+4x+7=5

The zeros of completesquare x^2+4x+7=5 are x=sqrt(2)-2,x=-sqrt(2)-2

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x^{2} + 4 x + 7 = 5

x = 2 - 2, x = - 2 - 2

Decimal

x = - 0.58578 \dots, x = - 3.41421 \dots

Solution steps

x^{2} + 4 x + 7 = 5

Move

7

to the right side

x^{2} + 4 x = - 2

Rewrite in the form

(x + a)^{2} = b

(x + 2)^{2} = 2

For

the solutions are

f (x) = a, - a

Solve

x + 2 = 2 : x = 2 - 2

Solve

x + 2 = - 2 : x = - 2 - 2

The solutions to the quadratic equation are:

x = 2 - 2, x = - 2 - 2

x^{2} - 6 x = \frac{1}{2} x + 42

5 n = 2 n^{2} - 3

2 x^{2} + 32 x + 126 = 0

\frac{x ( x - 3 )}{2} = x

x^{2} - 9 x - 20 = - 6 x - 2

What are the solutions to the equation completesquare x^2+4x+7=5 ?
The solutions to the equation completesquare x^2+4x+7=5 are x=sqrt(2)-2,x=-sqrt(2)-2
Find the zeros of completesquare x^2+4x+7=5
The zeros of completesquare x^2+4x+7=5 are x=sqrt(2)-2,x=-sqrt(2)-2