What are the solutions to the equation t^2-82t+160=0 ?

The solutions to the equation t^2-82t+160=0 are t=80,t=2

Find the zeros of t^2-82t+160=0

The zeros of t^2-82t+160=0 are t=80,t=2

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t^{2} - 82 t + 160 = 0

t = 80, t = 2

Solution steps

t^{2} - 82 t + 160 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 82 ) \pm ( - 82 ) ^{2} - 4 \cdot 1 \cdot 160}{2 \cdot 1}

(- 82)^{2} - 4 \cdot 1 \cdot 160 = 78

t_{1, 2} = \frac{- ( - 82 ) \pm 78}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 82 ) + 78}{2 \cdot 1}, t_{2} = \frac{- ( - 82 ) - 78}{2 \cdot 1}

t = \frac{- ( - 82 ) + 78}{2 \cdot 1} : 80

t = \frac{- ( - 82 ) - 78}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

t = 80, t = 2

9 w^{2} = 18 w + 7

z, z^{2} - 6 z + 25 = 0

x, x^{2} + y^{2} + 30 z - 50 = 0

x, t (x) = x^{2} - 1

3 x^{2} - 7 x + 2 = 5 x + 2

What are the solutions to the equation t^2-82t+160=0 ?
The solutions to the equation t^2-82t+160=0 are t=80,t=2
Find the zeros of t^2-82t+160=0
The zeros of t^2-82t+160=0 are t=80,t=2