What are the solutions to the equation 0=-5t^2+18t+16 ?

The solutions to the equation 0=-5t^2+18t+16 are t=-(-9+sqrt(161))/5 ,t=(9+sqrt(161))/5

Find the zeros of 0=-5t^2+18t+16

The zeros of 0=-5t^2+18t+16 are t=-(-9+sqrt(161))/5 ,t=(9+sqrt(161))/5

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0 = - 5 t^{2} + 18 t + 16

t = - \frac{- 9 + 161}{5}, t = \frac{9 + 161}{5}

Solution steps

0 = - 5 t^{2} + 18 t + 16

Switch sides

- 5 t^{2} + 18 t + 16 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 5 ) \cdot 16}{2 ( - 5 )}

1 8^{2} - 4 (- 5) \cdot 16 = 2161

t_{1, 2} = \frac{- 18 \pm 2 161}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 18 + 2 161}{2 ( - 5 )}, t_{2} = \frac{- 18 - 2 161}{2 ( - 5 )}

t = \frac{- 18 + 2 161}{2 ( - 5 )} : - \frac{- 9 + 161}{5}

t = \frac{- 18 - 2 161}{2 ( - 5 )} : \frac{9 + 161}{5}

The solutions to the quadratic equation are:

t = - \frac{- 9 + 161}{5}, t = \frac{9 + 161}{5}

(2 x + 1) + (3 x + 3) = 180

5 x^{2} - 5 x - 6 = 0

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

\frac{3}{8} + 2

8 x^{2} - 3 = 2

What are the solutions to the equation 0=-5t^2+18t+16 ?
The solutions to the equation 0=-5t^2+18t+16 are t=-(-9+sqrt(161))/5 ,t=(9+sqrt(161))/5
Find the zeros of 0=-5t^2+18t+16
The zeros of 0=-5t^2+18t+16 are t=-(-9+sqrt(161))/5 ,t=(9+sqrt(161))/5