What are the solutions to the equation t^2-5t-36=0 ?

The solutions to the equation t^2-5t-36=0 are t=9,t=-4

Find the zeros of t^2-5t-36=0

The zeros of t^2-5t-36=0 are t=9,t=-4

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t^{2} - 5 t - 36 = 0

t = 9, t = - 4

Solution steps

t^{2} - 5 t - 36 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 36) = 13

t_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 1}

t = \frac{- ( - 5 ) + 13}{2 \cdot 1} : 9

t = \frac{- ( - 5 ) - 13}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

t = 9, t = - 4

x^{2} + x^{2} = 4^{2}

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

x, x^{2} y + 5 x y + 4 y = 0

x^{2} - 14 x - 48 = 0

x^{2} + 4 x + 2

What are the solutions to the equation t^2-5t-36=0 ?
The solutions to the equation t^2-5t-36=0 are t=9,t=-4
Find the zeros of t^2-5t-36=0
The zeros of t^2-5t-36=0 are t=9,t=-4