What are the solutions to the equation 0=(x-5)^2+4(x-5)+4 ?

The solutions to the equation 0=(x-5)^2+4(x-5)+4 are x=3

Find the zeros of 0=(x-5)^2+4(x-5)+4

The zeros of 0=(x-5)^2+4(x-5)+4 are x=3

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0 = (x - 5)^{2} + 4 (x - 5) + 4

x = 3

Solution steps

0 = (x - 5)^{2} + 4 (x - 5) + 4

Expand

(x - 5)^{2} + 4 (x - 5) + 4 : x^{2} - 6 x + 9

0 = x^{2} - 6 x + 9

Switch sides

x^{2} - 6 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 9 = 0

x_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 \cdot 1}

x = \frac{- ( - 6 )}{2 \cdot 1}

\frac{- ( - 6 )}{2 \cdot 1} = 3

x = 3

The solution to the quadratic equation is:

x = 3

8 n^{2} - 3 = 37

5 x^{2} - 5 = x

m^{2} + 4 m = 32

5 x^{2} - 4 = x

y, s (y) = 3 yy + 5

What are the solutions to the equation 0=(x-5)^2+4(x-5)+4 ?
The solutions to the equation 0=(x-5)^2+4(x-5)+4 are x=3
Find the zeros of 0=(x-5)^2+4(x-5)+4
The zeros of 0=(x-5)^2+4(x-5)+4 are x=3