What are the solutions to the equation t^2+4t-140=0 ?

The solutions to the equation t^2+4t-140=0 are t=10,t=-14

Find the zeros of t^2+4t-140=0

The zeros of t^2+4t-140=0 are t=10,t=-14

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t^{2} + 4 t - 140 = 0

t = 10, t = - 14

Solution steps

t^{2} + 4 t - 140 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 140 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 140) = 24

t_{1, 2} = \frac{- 4 \pm 24}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 4 + 24}{2 \cdot 1}, t_{2} = \frac{- 4 - 24}{2 \cdot 1}

t = \frac{- 4 + 24}{2 \cdot 1} : 10

t = \frac{- 4 - 24}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

t = 10, t = - 14

(3 x + 1) (x - 1) = 0

13 x - 2 x^{2} = 0

- p^{2} + 2 p + 4 = 0

x^{2} + 24 = - 10 x

12 x - 6 x^{2} = 0

What are the solutions to the equation t^2+4t-140=0 ?
The solutions to the equation t^2+4t-140=0 are t=10,t=-14
Find the zeros of t^2+4t-140=0
The zeros of t^2+4t-140=0 are t=10,t=-14