What are the solutions to the equation 120=50t-5t^2 ?

The solutions to the equation 120=50t-5t^2 are t=4,t=6

Find the zeros of 120=50t-5t^2

The zeros of 120=50t-5t^2 are t=4,t=6

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120 = 50 t - 5 t^{2}

t = 4, t = 6

Solution steps

120 = 50 t - 5 t^{2}

Switch sides

50 t - 5 t^{2} = 120

Move

120

to the left side

50 t - 5 t^{2} - 120 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 5 t^{2} + 50 t - 120 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 ( - 5 ) ( - 120 )}{2 ( - 5 )}

5 0^{2} - 4 (- 5) (- 120) = 10

t_{1, 2} = \frac{- 50 \pm 10}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 50 + 10}{2 ( - 5 )}, t_{2} = \frac{- 50 - 10}{2 ( - 5 )}

t = \frac{- 50 + 10}{2 ( - 5 )} : 4

t = \frac{- 50 - 10}{2 ( - 5 )} : 6

The solutions to the quadratic equation are:

t = 4, t = 6

3 (x + 1)^{2} - 4 = 23

(2 x - 3)^{2} - (x + 5)^{2} = - 23

x^{2} + 1 x - 6 = 0

x^{2} - 16 x + 46 = 0

s^{2} + 4 s + 10

What are the solutions to the equation 120=50t-5t^2 ?
The solutions to the equation 120=50t-5t^2 are t=4,t=6
Find the zeros of 120=50t-5t^2
The zeros of 120=50t-5t^2 are t=4,t=6