What are the solutions to the equation t^2+t-6=0 ?

The solutions to the equation t^2+t-6=0 are t=2,t=-3

Find the zeros of t^2+t-6=0

The zeros of t^2+t-6=0 are t=2,t=-3

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t^{2} + t - 6 = 0

t = 2, t = - 3

Solution steps

t^{2} + t - 6 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 6) = 5

t_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 1 + 5}{2 \cdot 1}, t_{2} = \frac{- 1 - 5}{2 \cdot 1}

t = \frac{- 1 + 5}{2 \cdot 1} : 2

t = \frac{- 1 - 5}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

t = 2, t = - 3

6 - 2 x^{2} = 9 x + 15

x, 3 x^{2} y = 0

y, x^{2} + y^{2} - 2 z^{2} + z ln (x) = 0

x, x^{2} = 50

y^{2} - 6 y - 6 = 0

What are the solutions to the equation t^2+t-6=0 ?
The solutions to the equation t^2+t-6=0 are t=2,t=-3
Find the zeros of t^2+t-6=0
The zeros of t^2+t-6=0 are t=2,t=-3