What are the solutions to the equation 0=16t^2+60t-340 ?

The solutions to the equation 0=16t^2+60t-340 are t=(-15+sqrt(1585))/8 ,t=-(15+sqrt(1585))/8

Find the zeros of 0=16t^2+60t-340

The zeros of 0=16t^2+60t-340 are t=(-15+sqrt(1585))/8 ,t=-(15+sqrt(1585))/8

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0 = 16 t^{2} + 60 t - 340

t = \frac{- 15 + 1585}{8}, t = - \frac{15 + 1585}{8}

Decimal Notation

t = 3.10150 \dots, t = - 6.85150 \dots

Solution steps

0 = 16 t^{2} + 60 t - 340

Switch sides

16 t^{2} + 60 t - 340 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 \cdot 16 ( - 340 )}{2 \cdot 16}

6 0^{2} - 4 \cdot 16 (- 340) = 41585

t_{1, 2} = \frac{- 60 \pm 4 1585}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 60 + 4 1585}{2 \cdot 16}, t_{2} = \frac{- 60 - 4 1585}{2 \cdot 16}

t = \frac{- 60 + 4 1585}{2 \cdot 16} : \frac{- 15 + 1585}{8}

t = \frac{- 60 - 4 1585}{2 \cdot 16} : - \frac{15 + 1585}{8}

The solutions to the quadratic equation are:

t = \frac{- 15 + 1585}{8}, t = - \frac{15 + 1585}{8}

\frac{x + 3}{x - 2} < 2

2 x^{2} = 7 x + 6

\frac{1}{10}

x^{2} - 14 x = - 48

- 5 < x < 0

What are the solutions to the equation 0=16t^2+60t-340 ?
The solutions to the equation 0=16t^2+60t-340 are t=(-15+sqrt(1585))/8 ,t=-(15+sqrt(1585))/8
Find the zeros of 0=16t^2+60t-340
The zeros of 0=16t^2+60t-340 are t=(-15+sqrt(1585))/8 ,t=-(15+sqrt(1585))/8