What are the solutions to the equation t^2+7t-18=0 ?

The solutions to the equation t^2+7t-18=0 are t=2,t=-9

Find the zeros of t^2+7t-18=0

The zeros of t^2+7t-18=0 are t=2,t=-9

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t^{2} + 7 t - 18 = 0

t = 2, t = - 9

Solution steps

t^{2} + 7 t - 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 18) = 11

t_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 7 + 11}{2 \cdot 1}, t_{2} = \frac{- 7 - 11}{2 \cdot 1}

t = \frac{- 7 + 11}{2 \cdot 1} : 2

t = \frac{- 7 - 11}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

t = 2, t = - 9

- 2 x^{2} + 2 = - 3 x

7 (8 x + 3) (x - 7) = 0

9 p^{2} - 24 p + 14 = 0

(y - 2) (y - 3) = 9 y + 6

u^{2} - 11 u + 28 = 0

What are the solutions to the equation t^2+7t-18=0 ?
The solutions to the equation t^2+7t-18=0 are t=2,t=-9
Find the zeros of t^2+7t-18=0
The zeros of t^2+7t-18=0 are t=2,t=-9