What are the solutions to the equation 918=5t^2+3t+280 ?

The solutions to the equation 918=5t^2+3t+280 are t=11,t=-58/5

Find the zeros of 918=5t^2+3t+280

The zeros of 918=5t^2+3t+280 are t=11,t=-58/5

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918 = 5 t^{2} + 3 t + 280

t = 11, t = - \frac{58}{5}

Decimal

t = 11, t = - 11.6

Solution steps

918 = 5 t^{2} + 3 t + 280

Switch sides

5 t^{2} + 3 t + 280 = 918

Move

918

to the left side

5 t^{2} + 3 t - 638 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 638 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 638) = 113

t_{1, 2} = \frac{- 3 \pm 113}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- 3 + 113}{2 \cdot 5}, t_{2} = \frac{- 3 - 113}{2 \cdot 5}

t = \frac{- 3 + 113}{2 \cdot 5} : 11

t = \frac{- 3 - 113}{2 \cdot 5} : - \frac{58}{5}

The solutions to the quadratic equation are:

t = 11, t = - \frac{58}{5}

25 n^{2} - 24 n + 6 = n

5 x^{2} - 12 x = 2 x^{2}

16 (4 - 2 x)^{2} = 9

(2 x + 3)^{2} = 3 x (x + 2)

22 c^{2} - 35 c + 12 = - 3 c^{2}

What are the solutions to the equation 918=5t^2+3t+280 ?
The solutions to the equation 918=5t^2+3t+280 are t=11,t=-58/5
Find the zeros of 918=5t^2+3t+280
The zeros of 918=5t^2+3t+280 are t=11,t=-58/5