What are the solutions to the equation t^2+3t-18=0 ?

The solutions to the equation t^2+3t-18=0 are t=3,t=-6

Find the zeros of t^2+3t-18=0

The zeros of t^2+3t-18=0 are t=3,t=-6

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t^{2} + 3 t - 18 = 0

t = 3, t = - 6

Solution steps

t^{2} + 3 t - 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 18) = 9

t_{1, 2} = \frac{- 3 \pm 9}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 9}{2 \cdot 1}, t_{2} = \frac{- 3 - 9}{2 \cdot 1}

t = \frac{- 3 + 9}{2 \cdot 1} : 3

t = \frac{- 3 - 9}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

t = 3, t = - 6

\frac{x - 4}{2} = \frac{2 x + 1}{3}

\frac{e ^{7 i} - e ^{- 7 i}}{2 i}

1 - \frac{1}{6}

5 x^{2} + 9 x + 2 = 0

24 = 1.25 (1 + \frac{y}{1})

What are the solutions to the equation t^2+3t-18=0 ?
The solutions to the equation t^2+3t-18=0 are t=3,t=-6
Find the zeros of t^2+3t-18=0
The zeros of t^2+3t-18=0 are t=3,t=-6