What are the solutions to the equation 65=4.5t^2+43.5t+17 ?

The solutions to the equation 65=4.5t^2+43.5t+17 are t=1,t=-32/3

Find the zeros of 65=4.5t^2+43.5t+17

The zeros of 65=4.5t^2+43.5t+17 are t=1,t=-32/3

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65 = 4.5 t^{2} + 43.5 t + 17

t = 1, t = - \frac{32}{3}

Decimal Notation

t = 1, t = - 10.66666 \dots

Solution steps

65 = 4.5 t^{2} + 43.5 t + 17

Multiply both sides by

10

650 = 45 t^{2} + 435 t + 170

Switch sides

45 t^{2} + 435 t + 170 = 650

Move

650

to the left side

45 t^{2} + 435 t - 480 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 435 \pm 43 5 ^{2} - 4 \cdot 45 ( - 480 )}{2 \cdot 45}

43 5^{2} - 4 \cdot 45 (- 480) = 525

t_{1, 2} = \frac{- 435 \pm 525}{2 \cdot 45}

Separate the solutions

t_{1} = \frac{- 435 + 525}{2 \cdot 45}, t_{2} = \frac{- 435 - 525}{2 \cdot 45}

t = \frac{- 435 + 525}{2 \cdot 45} : 1

t = \frac{- 435 - 525}{2 \cdot 45} : - \frac{32}{3}

The solutions to the quadratic equation are:

t = 1, t = - \frac{32}{3}

15 + 4 y^{2} = 17 y

12 x^{2} + 18 x - 12 = 0

(5 b + 1) (6 b + 1) = 0

t, y = - 14 t^{2} + V o t

36 x^{2} - 396 x + 1080 = 0

What are the solutions to the equation 65=4.5t^2+43.5t+17 ?
The solutions to the equation 65=4.5t^2+43.5t+17 are t=1,t=-32/3
Find the zeros of 65=4.5t^2+43.5t+17
The zeros of 65=4.5t^2+43.5t+17 are t=1,t=-32/3