What are the solutions to the equation 40=30t-5(t)^2 ?

The solutions to the equation 40=30t-5(t)^2 are t=2,t=4

Find the zeros of 40=30t-5(t)^2

The zeros of 40=30t-5(t)^2 are t=2,t=4

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40 = 30 t - 5 (t)^{2}

t = 2, t = 4

Solution steps

40 = 30 t - 5 t^{2}

Switch sides

30 t - 5 t^{2} = 40

Move

40

to the left side

30 t - 5 t^{2} - 40 = 0

Write in the standard form

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

- 5 t^{2} + 30 t - 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 ( - 5 ) ( - 40 )}{2 ( - 5 )}

3 0^{2} - 4 (- 5) (- 40) = 10

t_{1, 2} = \frac{- 30 \pm 10}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 30 + 10}{2 ( - 5 )}, t_{2} = \frac{- 30 - 10}{2 ( - 5 )}

t = \frac{- 30 + 10}{2 ( - 5 )} : 2

t = \frac{- 30 - 10}{2 ( - 5 )} : 4

The solutions to the quadratic equation are:

t = 2, t = 4

4 x^{2} - 9 x = 28

x^{2} + 3 x - 208 = 0

y, 2 x^{2} + 2 x y + 4 y^{2} = 5

2 x^{2} + 5 x - 3

(x - 1) (x + 2) = e

What are the solutions to the equation 40=30t-5(t)^2 ?
The solutions to the equation 40=30t-5(t)^2 are t=2,t=4
Find the zeros of 40=30t-5(t)^2
The zeros of 40=30t-5(t)^2 are t=2,t=4