What are the solutions to the equation 136=16t^2+100t ?

The solutions to the equation 136=16t^2+100t are t=(-25+sqrt(1169))/8 ,t=-(25+sqrt(1169))/8

Find the zeros of 136=16t^2+100t

The zeros of 136=16t^2+100t are t=(-25+sqrt(1169))/8 ,t=-(25+sqrt(1169))/8

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136 = 16 t^{2} + 100 t

t = \frac{- 25 + 1169}{8}, t = - \frac{25 + 1169}{8}

Decimal Notation

t = 1.14883 \dots, t = - 7.39883 \dots

Solution steps

136 = 16 t^{2} + 100 t

Switch sides

16 t^{2} + 100 t = 136

Move

136

to the left side

16 t^{2} + 100 t - 136 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 16 ( - 136 )}{2 \cdot 16}

10 0^{2} - 4 \cdot 16 (- 136) = 41169

t_{1, 2} = \frac{- 100 \pm 4 1169}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 100 + 4 1169}{2 \cdot 16}, t_{2} = \frac{- 100 - 4 1169}{2 \cdot 16}

t = \frac{- 100 + 4 1169}{2 \cdot 16} : \frac{- 25 + 1169}{8}

t = \frac{- 100 - 4 1169}{2 \cdot 16} : - \frac{25 + 1169}{8}

The solutions to the quadratic equation are:

t = \frac{- 25 + 1169}{8}, t = - \frac{25 + 1169}{8}

23 y + 7 y^{2} = 20

2 - t - t^{2} = 0

- 3 = 4 (x)^{2} - (x) - 3

(c - 3)^{2} = 100

V, q = \frac{1}{2} p V^{2}

What are the solutions to the equation 136=16t^2+100t ?
The solutions to the equation 136=16t^2+100t are t=(-25+sqrt(1169))/8 ,t=-(25+sqrt(1169))/8
Find the zeros of 136=16t^2+100t
The zeros of 136=16t^2+100t are t=(-25+sqrt(1169))/8 ,t=-(25+sqrt(1169))/8