What are the solutions to the equation u^2-8u+7=0 ?

The solutions to the equation u^2-8u+7=0 are u=7,u=1

Find the zeros of u^2-8u+7=0

The zeros of u^2-8u+7=0 are u=7,u=1

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u^{2} - 8 u + 7 = 0

u = 7, u = 1

Solution steps

u^{2} - 8 u + 7 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 7 = 6

u_{1, 2} = \frac{- ( - 8 ) \pm 6}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 6}{2 \cdot 1}, u_{2} = \frac{- ( - 8 ) - 6}{2 \cdot 1}

u = \frac{- ( - 8 ) + 6}{2 \cdot 1} : 7

u = \frac{- ( - 8 ) - 6}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

u = 7, u = 1

x^{2} - 3 x + 54 = 0

t, r (t) = a \frac{t ^{2}}{2}

y^{2} = 1 5^{2} - 9^{2}

4 x^{2} - 3 - 4 x = 0

y, 2 x^{2} = y^{2}

What are the solutions to the equation u^2-8u+7=0 ?
The solutions to the equation u^2-8u+7=0 are u=7,u=1
Find the zeros of u^2-8u+7=0
The zeros of u^2-8u+7=0 are u=7,u=1