What are the solutions to the equation 0=130-6t-16t^2 ?

The solutions to the equation 0=130-6t-16t^2 are t=-(3+sqrt(2089))/(16),t=(sqrt(2089)-3)/(16)

Find the zeros of 0=130-6t-16t^2

The zeros of 0=130-6t-16t^2 are t=-(3+sqrt(2089))/(16),t=(sqrt(2089)-3)/(16)

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0 = 130 - 6 t - 16 t^{2}

t = - \frac{3 + 2089}{16}, t = \frac{2089 - 3}{16}

Solution steps

0 = 130 - 6 t - 16 t^{2}

Switch sides

130 - 6 t - 16 t^{2} = 0

Write in the standard form

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

- 16 t^{2} - 6 t + 130 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 16 ) \cdot 130}{2 ( - 16 )}

(- 6)^{2} - 4 (- 16) \cdot 130 = 22089

t_{1, 2} = \frac{- ( - 6 ) \pm 2 2089}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 6 ) + 2 2089}{2 ( - 16 )}, t_{2} = \frac{- ( - 6 ) - 2 2089}{2 ( - 16 )}

t = \frac{- ( - 6 ) + 2 2089}{2 ( - 16 )} : - \frac{3 + 2089}{16}

t = \frac{- ( - 6 ) - 2 2089}{2 ( - 16 )} : \frac{2089 - 3}{16}

The solutions to the quadratic equation are:

t = - \frac{3 + 2089}{16}, t = \frac{2089 - 3}{16}

n^{2} - 4 = 4

w \frac{5 w + 4}{3} = \frac{3 w}{2}

(k + 4)^{2} = 25

x, f (x - 3) = x^{2} + 9 x + 5

104 = u^{2} - 5 u

What are the solutions to the equation 0=130-6t-16t^2 ?
The solutions to the equation 0=130-6t-16t^2 are t=-(3+sqrt(2089))/(16),t=(sqrt(2089)-3)/(16)
Find the zeros of 0=130-6t-16t^2
The zeros of 0=130-6t-16t^2 are t=-(3+sqrt(2089))/(16),t=(sqrt(2089)-3)/(16)