What are the solutions to the equation 0=-16t^2-50t+240 ?

The solutions to the equation 0=-16t^2-50t+240 are t=-(25+sqrt(4465))/(16),t=(sqrt(4465)-25)/(16)

Find the zeros of 0=-16t^2-50t+240

The zeros of 0=-16t^2-50t+240 are t=-(25+sqrt(4465))/(16),t=(sqrt(4465)-25)/(16)

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0 = - 16 t^{2} - 50 t + 240

t = - \frac{25 + 4465}{16}, t = \frac{4465 - 25}{16}

Decimal Notation

t = - 5.73879 \dots, t = 2.61379 \dots

Solution steps

0 = - 16 t^{2} - 50 t + 240

Switch sides

- 16 t^{2} - 50 t + 240 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 50 ) \pm ( - 50 ) ^{2} - 4 ( - 16 ) \cdot 240}{2 ( - 16 )}

(- 50)^{2} - 4 (- 16) \cdot 240 = 24465

t_{1, 2} = \frac{- ( - 50 ) \pm 2 4465}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 50 ) + 2 4465}{2 ( - 16 )}, t_{2} = \frac{- ( - 50 ) - 2 4465}{2 ( - 16 )}

t = \frac{- ( - 50 ) + 2 4465}{2 ( - 16 )} : - \frac{25 + 4465}{16}

t = \frac{- ( - 50 ) - 2 4465}{2 ( - 16 )} : \frac{4465 - 25}{16}

The solutions to the quadratic equation are:

t = - \frac{25 + 4465}{16}, t = \frac{4465 - 25}{16}

12 5^{\frac{2}{3}}

m^{2} + 4 mn + 4 n^{2}

3 lo g_{b} (x) - (lo g_{b} (4) + lo g_{b} (x))

∣ x ∣ + ∣ x + 1 ∣ = 0

∣ 2 x - 1 ∣ + 5 = 6

What are the solutions to the equation 0=-16t^2-50t+240 ?
The solutions to the equation 0=-16t^2-50t+240 are t=-(25+sqrt(4465))/(16),t=(sqrt(4465)-25)/(16)
Find the zeros of 0=-16t^2-50t+240
The zeros of 0=-16t^2-50t+240 are t=-(25+sqrt(4465))/(16),t=(sqrt(4465)-25)/(16)