What are the solutions to the equation 72=-16(t)^2+28(t)+216 ?

The solutions to the equation 72=-16(t)^2+28(t)+216 are t=-9/4 ,t=4

Find the zeros of 72=-16(t)^2+28(t)+216

The zeros of 72=-16(t)^2+28(t)+216 are t=-9/4 ,t=4

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72 = - 16 (t)^{2} + 28 (t) + 216

t = - \frac{9}{4}, t = 4

Decimal Notation

t = - 2.25, t = 4

Solution steps

72 = - 16 t^{2} + 28 (t) + 216

Refine

72 = - 16 t^{2} + 28 t + 216

Switch sides

- 16 t^{2} + 28 t + 216 = 72

Move

72

to the left side

- 16 t^{2} + 28 t + 144 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 ( - 16 ) \cdot 144}{2 ( - 16 )}

2 8^{2} - 4 (- 16) \cdot 144 = 100

t_{1, 2} = \frac{- 28 \pm 100}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 28 + 100}{2 ( - 16 )}, t_{2} = \frac{- 28 - 100}{2 ( - 16 )}

t = \frac{- 28 + 100}{2 ( - 16 )} : - \frac{9}{4}

t = \frac{- 28 - 100}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

t = - \frac{9}{4}, t = 4

\frac{x + 1}{x - 4} > 0

12 p^{2} + 32 p - 4 = 8

5 x^{2} - 30 x = 5

(x + 2) (x^{2} - 12 x + 35)

100 x - x^{3} \geq 0

What are the solutions to the equation 72=-16(t)^2+28(t)+216 ?
The solutions to the equation 72=-16(t)^2+28(t)+216 are t=-9/4 ,t=4
Find the zeros of 72=-16(t)^2+28(t)+216
The zeros of 72=-16(t)^2+28(t)+216 are t=-9/4 ,t=4