What are the solutions to the equation x^2+4x+4=49 ?

The solutions to the equation x^2+4x+4=49 are x=5,x=-9

Find the zeros of x^2+4x+4=49

The zeros of x^2+4x+4=49 are x=5,x=-9

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x^{2} + 4 x + 4 = 49

x = 5, x = - 9

Solution steps

x^{2} + 4 x + 4 = 49

Move

49

to the left side

x^{2} + 4 x - 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 45 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 45) = 14

x_{1, 2} = \frac{- 4 \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 14}{2 \cdot 1}, x_{2} = \frac{- 4 - 14}{2 \cdot 1}

x = \frac{- 4 + 14}{2 \cdot 1} : 5

x = \frac{- 4 - 14}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = 5, x = - 9

x (2 x - 13) = 15

f = 2 + 5

\frac{6 !}{1 ! ( 6 - 1 ) !}

14 \cdot 7

\frac{1 - 3 i}{2 + i}

What are the solutions to the equation x^2+4x+4=49 ?
The solutions to the equation x^2+4x+4=49 are x=5,x=-9
Find the zeros of x^2+4x+4=49
The zeros of x^2+4x+4=49 are x=5,x=-9