What are the solutions to the equation t^2-25t+144=0 ?

The solutions to the equation t^2-25t+144=0 are t=16,t=9

Find the zeros of t^2-25t+144=0

The zeros of t^2-25t+144=0 are t=16,t=9

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t^{2} - 25 t + 144 = 0

t = 16, t = 9

Solution steps

t^{2} - 25 t + 144 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 25 ) \pm ( - 25 ) ^{2} - 4 \cdot 1 \cdot 144}{2 \cdot 1}

(- 25)^{2} - 4 \cdot 1 \cdot 144 = 7

t_{1, 2} = \frac{- ( - 25 ) \pm 7}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 25 ) + 7}{2 \cdot 1}, t_{2} = \frac{- ( - 25 ) - 7}{2 \cdot 1}

t = \frac{- ( - 25 ) + 7}{2 \cdot 1} : 16

t = \frac{- ( - 25 ) - 7}{2 \cdot 1} : 9

The solutions to the quadratic equation are:

t = 16, t = 9

(w + 6)^{2} - 40 = 0

x^{2} - 16 x + 13 = 0

(6 x - 1)^{2} = 43

x^{2} - 8 x + 47 = - 5

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

What are the solutions to the equation t^2-25t+144=0 ?
The solutions to the equation t^2-25t+144=0 are t=16,t=9
Find the zeros of t^2-25t+144=0
The zeros of t^2-25t+144=0 are t=16,t=9