What are the solutions to the equation t^2-4t-12=0 ?

The solutions to the equation t^2-4t-12=0 are t=6,t=-2

Find the zeros of t^2-4t-12=0

The zeros of t^2-4t-12=0 are t=6,t=-2

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t^{2} - 4 t - 12 = 0

t = 6, t = - 2

Solution steps

t^{2} - 4 t - 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

t_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, t_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

t = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

t = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

t = 6, t = - 2

x, x^{2} y + x y^{2} = 3 x

x^{2} - 8 = 8

r^{2} - 8 r = - 8

12 x^{2} - 17 x + 6 = 0

- 3 x^{2} - 39 x + 133 = 9 x + 1

What are the solutions to the equation t^2-4t-12=0 ?
The solutions to the equation t^2-4t-12=0 are t=6,t=-2
Find the zeros of t^2-4t-12=0
The zeros of t^2-4t-12=0 are t=6,t=-2