What are the solutions to the equation (t+3)^2-(t+3)-12=0 ?

The solutions to the equation (t+3)^2-(t+3)-12=0 are t=1,t=-6

Find the zeros of (t+3)^2-(t+3)-12=0

The zeros of (t+3)^2-(t+3)-12=0 are t=1,t=-6

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(t + 3)^{2} - (t + 3) - 12 = 0

t = 1, t = - 6

Solution steps

(t + 3)^{2} - (t + 3) - 12 = 0

Expand

(t + 3)^{2} - (t + 3) - 12 : t^{2} + 5 t - 6

t^{2} + 5 t - 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 6) = 7

t_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 5 + 7}{2 \cdot 1}, t_{2} = \frac{- 5 - 7}{2 \cdot 1}

t = \frac{- 5 + 7}{2 \cdot 1} : 1

t = \frac{- 5 - 7}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

t = 1, t = - 6

t^{2} - 2 t + 1

\frac{3}{4} \cdot \frac{4}{9} \cdot \frac{2}{6}

(2 - 3)^{3}

\frac{\frac{1}{2}}{850}

x^{2} + 13 x - 27 = 0

What are the solutions to the equation (t+3)^2-(t+3)-12=0 ?
The solutions to the equation (t+3)^2-(t+3)-12=0 are t=1,t=-6
Find the zeros of (t+3)^2-(t+3)-12=0
The zeros of (t+3)^2-(t+3)-12=0 are t=1,t=-6