What are the solutions to the equation 1009=-2t^2+108t+1 ?

The solutions to the equation 1009=-2t^2+108t+1 are t=12,t=42

Find the zeros of 1009=-2t^2+108t+1

The zeros of 1009=-2t^2+108t+1 are t=12,t=42

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1009 = - 2 t^{2} + 108 t + 1

t = 12, t = 42

Solution steps

1009 = - 2 t^{2} + 108 t + 1

Switch sides

- 2 t^{2} + 108 t + 1 = 1009

Move

1009

to the left side

- 2 t^{2} + 108 t - 1008 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 108 \pm 10 8 ^{2} - 4 ( - 2 ) ( - 1008 )}{2 ( - 2 )}

10 8^{2} - 4 (- 2) (- 1008) = 60

t_{1, 2} = \frac{- 108 \pm 60}{2 ( - 2 )}

Separate the solutions

t_{1} = \frac{- 108 + 60}{2 ( - 2 )}, t_{2} = \frac{- 108 - 60}{2 ( - 2 )}

t = \frac{- 108 + 60}{2 ( - 2 )} : 12

t = \frac{- 108 - 60}{2 ( - 2 )} : 42

The solutions to the quadratic equation are:

t = 12, t = 42

5.6 \cdot 1 0^{- 10} = \frac{x ^{2}}{0.5}

j (t) = - \frac{1}{3} t^{2} + 6 t - 10

x, x^{2} - 2 x y - y^{2} = 6

(x - 11) (3 x + 9) = 0

v^{2} - 15 = - 2 v

What are the solutions to the equation 1009=-2t^2+108t+1 ?
The solutions to the equation 1009=-2t^2+108t+1 are t=12,t=42
Find the zeros of 1009=-2t^2+108t+1
The zeros of 1009=-2t^2+108t+1 are t=12,t=42