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

The solutions to the equation t^2+t-2=0 are t=1,t=-2

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

The zeros of t^2+t-2=0 are t=1,t=-2

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t^{2} + t - 2 = 0

t = 1, t = - 2

Solution steps

t^{2} + t - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

t_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 1 + 3}{2 \cdot 1}, t_{2} = \frac{- 1 - 3}{2 \cdot 1}

t = \frac{- 1 + 3}{2 \cdot 1} : 1

t = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

t = 1, t = - 2

x^{2} - 2 x - 8 = 0

(n - 6) (n - 2) = - 9 n + 32

3 x^{2} - 16 x + 24 = 0

p, p^{1} \frac{v ^{1}}{t ^{1}} = p^{2} \frac{v ^{2}}{t ^{2}}

(x + 1) - x (x - 3) = 6

What are the solutions to the equation t^2+t-2=0 ?
The solutions to the equation t^2+t-2=0 are t=1,t=-2
Find the zeros of t^2+t-2=0
The zeros of t^2+t-2=0 are t=1,t=-2