What are the solutions to the equation u^2-8u+16=0 ?

The solutions to the equation u^2-8u+16=0 are u=4

Find the zeros of u^2-8u+16=0

The zeros of u^2-8u+16=0 are u=4

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u^{2} - 8 u + 16 = 0

u = 4

Solution steps

u^{2} - 8 u + 16 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 16 = 0

u_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 1}

u = \frac{- ( - 8 )}{2 \cdot 1}

\frac{- ( - 8 )}{2 \cdot 1} = 4

u = 4

The solution to the quadratic equation is:

u = 4

x, x^{2} + y^{2} + z^{2} = 36

3 x^{2} + 14 x - 24 = 0

x^{2} + 4 x = 0

2 x^{2} + 4 x - 6 = 0

7 x^{2} + 23 x + 168 = 6 x^{2} + 3 x + 7

What are the solutions to the equation u^2-8u+16=0 ?
The solutions to the equation u^2-8u+16=0 are u=4
Find the zeros of u^2-8u+16=0
The zeros of u^2-8u+16=0 are u=4