What are the solutions to the equation 15t-1.86t^2-25=0 ?

The solutions to the equation 15t-1.86t^2-25=0 are t=(25(15-sqrt(39)))/(93),t=(25(15+sqrt(39)))/(93)

Find the zeros of 15t-1.86t^2-25=0

The zeros of 15t-1.86t^2-25=0 are t=(25(15-sqrt(39)))/(93),t=(25(15+sqrt(39)))/(93)

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15 t - 1.86 t^{2} - 25 = 0

t = \frac{25 ( 15 - 39 )}{93}, t = \frac{25 ( 15 + 39 )}{93}

Decimal Notation

t = 2.35349 \dots, t = 5.71102 \dots

Solution steps

15 t - 1.86 t^{2} - 25 = 0

Multiply both sides by

100

1500 t - 186 t^{2} - 2500 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 186 t^{2} + 1500 t - 2500 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 1500 \pm 150 0 ^{2} - 4 ( - 186 ) ( - 2500 )}{2 ( - 186 )}

150 0^{2} - 4 (- 186) (- 2500) = 10039

t_{1, 2} = \frac{- 1500 \pm 100 39}{2 ( - 186 )}

Separate the solutions

t_{1} = \frac{- 1500 + 100 39}{2 ( - 186 )}, t_{2} = \frac{- 1500 - 100 39}{2 ( - 186 )}

t = \frac{- 1500 + 100 39}{2 ( - 186 )} : \frac{25 ( 15 - 39 )}{93}

t = \frac{- 1500 - 100 39}{2 ( - 186 )} : \frac{25 ( 15 + 39 )}{93}

The solutions to the quadratic equation are:

t = \frac{25 ( 15 - 39 )}{93}, t = \frac{25 ( 15 + 39 )}{93}

x^{2} - 4 x = 2 x - x^{2}

56 = - x^{2} + 15 x

(x - 1) (5 x + 1) = 216

- 0.5 x^{2} - 30 x + 350 = 0

2.5 x^{2} = 5 x + 50 - 8487.05 - 848.705 x

What are the solutions to the equation 15t-1.86t^2-25=0 ?
The solutions to the equation 15t-1.86t^2-25=0 are t=(25(15-sqrt(39)))/(93),t=(25(15+sqrt(39)))/(93)
Find the zeros of 15t-1.86t^2-25=0
The zeros of 15t-1.86t^2-25=0 are t=(25(15-sqrt(39)))/(93),t=(25(15+sqrt(39)))/(93)