What are the solutions to the equation 28=-16(t)^2+34(t)+148 ?

The solutions to the equation 28=-16(t)^2+34(t)+148 are t=-15/8 ,t=4

Find the zeros of 28=-16(t)^2+34(t)+148

The zeros of 28=-16(t)^2+34(t)+148 are t=-15/8 ,t=4

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28 = - 16 (t)^{2} + 34 (t) + 148

t = - \frac{15}{8}, t = 4

Decimal Notation

t = - 1.875, t = 4

Solution steps

28 = - 16 t^{2} + 34 (t) + 148

Refine

28 = - 16 t^{2} + 34 t + 148

Switch sides

- 16 t^{2} + 34 t + 148 = 28

Move

28

to the left side

- 16 t^{2} + 34 t + 120 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 34 \pm 3 4 ^{2} - 4 ( - 16 ) \cdot 120}{2 ( - 16 )}

3 4^{2} - 4 (- 16) \cdot 120 = 94

t_{1, 2} = \frac{- 34 \pm 94}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 34 + 94}{2 ( - 16 )}, t_{2} = \frac{- 34 - 94}{2 ( - 16 )}

t = \frac{- 34 + 94}{2 ( - 16 )} : - \frac{15}{8}

t = \frac{- 34 - 94}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = - \frac{15}{8}, t = 4

5 (x - 7) = 6 (x + 2)

- 6 p - p = 14

169 x^{2} - 64

6 p + 16 p^{2} + 115 = 0

\frac{3 ^{- 2}}{9}

What are the solutions to the equation 28=-16(t)^2+34(t)+148 ?
The solutions to the equation 28=-16(t)^2+34(t)+148 are t=-15/8 ,t=4
Find the zeros of 28=-16(t)^2+34(t)+148
The zeros of 28=-16(t)^2+34(t)+148 are t=-15/8 ,t=4