What are the solutions to the equation 20=-1.9t^2+18t+1 ?

The solutions to the equation 20=-1.9t^2+18t+1 are t=(90-sqrt(4490))/(19),t=(90+sqrt(4490))/(19)

Find the zeros of 20=-1.9t^2+18t+1

The zeros of 20=-1.9t^2+18t+1 are t=(90-sqrt(4490))/(19),t=(90+sqrt(4490))/(19)

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20 = - 1.9 t^{2} + 18 t + 1

t = \frac{90 - 4490}{19}, t = \frac{90 + 4490}{19}

Decimal Notation

t = 1.21013 \dots, t = 8.26355 \dots

Solution steps

20 = - 1.9 t^{2} + 18 t + 1

Multiply both sides by

10

200 = - 19 t^{2} + 180 t + 10

Switch sides

- 19 t^{2} + 180 t + 10 = 200

Move

200

to the left side

- 19 t^{2} + 180 t - 190 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 180 \pm 18 0 ^{2} - 4 ( - 19 ) ( - 190 )}{2 ( - 19 )}

18 0^{2} - 4 (- 19) (- 190) = 24490

t_{1, 2} = \frac{- 180 \pm 2 4490}{2 ( - 19 )}

Separate the solutions

t_{1} = \frac{- 180 + 2 4490}{2 ( - 19 )}, t_{2} = \frac{- 180 - 2 4490}{2 ( - 19 )}

t = \frac{- 180 + 2 4490}{2 ( - 19 )} : \frac{90 - 4490}{19}

t = \frac{- 180 - 2 4490}{2 ( - 19 )} : \frac{90 + 4490}{19}

The solutions to the quadratic equation are:

t = \frac{90 - 4490}{19}, t = \frac{90 + 4490}{19}

x - 2 = 10 - x^{2}

- x^{2} - 10 x - 9 = 0

s^{2} + 10 s + 25 = 0

4 z^{2} = 16 z

6 x^{2} - 11 x + 6 = 0

What are the solutions to the equation 20=-1.9t^2+18t+1 ?
The solutions to the equation 20=-1.9t^2+18t+1 are t=(90-sqrt(4490))/(19),t=(90+sqrt(4490))/(19)
Find the zeros of 20=-1.9t^2+18t+1
The zeros of 20=-1.9t^2+18t+1 are t=(90-sqrt(4490))/(19),t=(90+sqrt(4490))/(19)