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

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

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

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

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

t = 9, t = - 2

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} : 9

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

The solutions to the quadratic equation are:

t = 9, t = - 2

\frac{- 5. 5 ^{2}}{5} + 2 \cdot 5.5

47 \leq 8 - 7 t

1.5 \cdot 2

\frac{15 c}{12 ( d - 3 )} \cdot \frac{21 d}{6 c}

8 x^{3} - 64 y^{3}

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