What are the solutions to the equation 0=16t^2+40t-280 ?

The solutions to the equation 0=16t^2+40t-280 are t=(-5+sqrt(305))/4 ,t=-(5+sqrt(305))/4

Find the zeros of 0=16t^2+40t-280

The zeros of 0=16t^2+40t-280 are t=(-5+sqrt(305))/4 ,t=-(5+sqrt(305))/4

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0 = 16 t^{2} + 40 t - 280

t = \frac{- 5 + 305}{4}, t = - \frac{5 + 305}{4}

Decimal Notation

t = 3.11606 \dots, t = - 5.61606 \dots

Solution steps

0 = 16 t^{2} + 40 t - 280

Switch sides

16 t^{2} + 40 t - 280 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 16 ( - 280 )}{2 \cdot 16}

4 0^{2} - 4 \cdot 16 (- 280) = 8305

t_{1, 2} = \frac{- 40 \pm 8 305}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 40 + 8 305}{2 \cdot 16}, t_{2} = \frac{- 40 - 8 305}{2 \cdot 16}

t = \frac{- 40 + 8 305}{2 \cdot 16} : \frac{- 5 + 305}{4}

t = \frac{- 40 - 8 305}{2 \cdot 16} : - \frac{5 + 305}{4}

The solutions to the quadratic equation are:

t = \frac{- 5 + 305}{4}, t = - \frac{5 + 305}{4}

∣ 2 x + 5 ∣ < 4

x^{2} z^{2} + 36 y^{2} - 4 y^{2} z^{2} - 9 x^{2}

z^{0}

- 4 (u + 2) + 2 = 2 (u + 6)

x^{2} - 11 x + 5 = - 6 x + 5

What are the solutions to the equation 0=16t^2+40t-280 ?
The solutions to the equation 0=16t^2+40t-280 are t=(-5+sqrt(305))/4 ,t=-(5+sqrt(305))/4
Find the zeros of 0=16t^2+40t-280
The zeros of 0=16t^2+40t-280 are t=(-5+sqrt(305))/4 ,t=-(5+sqrt(305))/4