What are the solutions to the equation 115=-15t^2+120t+10 ?

The solutions to the equation 115=-15t^2+120t+10 are t=1,t=7

Find the zeros of 115=-15t^2+120t+10

The zeros of 115=-15t^2+120t+10 are t=1,t=7

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115 = - 15 t^{2} + 120 t + 10

t = 1, t = 7

Solution steps

115 = - 15 t^{2} + 120 t + 10

Switch sides

- 15 t^{2} + 120 t + 10 = 115

Move

115

to the left side

- 15 t^{2} + 120 t - 105 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 15 ) ( - 105 )}{2 ( - 15 )}

12 0^{2} - 4 (- 15) (- 105) = 90

t_{1, 2} = \frac{- 120 \pm 90}{2 ( - 15 )}

Separate the solutions

t_{1} = \frac{- 120 + 90}{2 ( - 15 )}, t_{2} = \frac{- 120 - 90}{2 ( - 15 )}

t = \frac{- 120 + 90}{2 ( - 15 )} : 1

t = \frac{- 120 - 90}{2 ( - 15 )} : 7

The solutions to the quadratic equation are:

t = 1, t = 7

a^{2} - 4 a

(- \frac{1}{3})^{3}

2 (x + 7)^{2} = 36

e^{- i (\frac{π}{2})}

What are the solutions to the equation 115=-15t^2+120t+10 ?
The solutions to the equation 115=-15t^2+120t+10 are t=1,t=7
Find the zeros of 115=-15t^2+120t+10
The zeros of 115=-15t^2+120t+10 are t=1,t=7