What are the solutions to the equation t^2-9t+20=0 ?

The solutions to the equation t^2-9t+20=0 are t=5,t=4

Find the zeros of t^2-9t+20=0

The zeros of t^2-9t+20=0 are t=5,t=4

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t^{2} - 9 t + 20 = 0

t = 5, t = 4

Solution steps

t^{2} - 9 t + 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 20 = 1

t_{1, 2} = \frac{- ( - 9 ) \pm 1}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 9 ) + 1}{2 \cdot 1}, t_{2} = \frac{- ( - 9 ) - 1}{2 \cdot 1}

t = \frac{- ( - 9 ) + 1}{2 \cdot 1} : 5

t = \frac{- ( - 9 ) - 1}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

t = 5, t = 4

0 = 3 x^{2} - 6 x + 7

1600 = 2 y^{2}

1 - 4 x - 4 x^{2} = 0

2 x (x - 3) = x + 4

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

What are the solutions to the equation t^2-9t+20=0 ?
The solutions to the equation t^2-9t+20=0 are t=5,t=4
Find the zeros of t^2-9t+20=0
The zeros of t^2-9t+20=0 are t=5,t=4