What are the solutions to the equation t^2-2t(t+1)-8(t+1)^2=0 ?

The solutions to the equation t^2-2t(t+1)-8(t+1)^2=0 are t=-4/3 ,t=-2/3

Find the zeros of t^2-2t(t+1)-8(t+1)^2=0

The zeros of t^2-2t(t+1)-8(t+1)^2=0 are t=-4/3 ,t=-2/3

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t^{2} - 2 t (t + 1) - 8 (t + 1)^{2} = 0

t = - \frac{4}{3}, t = - \frac{2}{3}

Decimal Notation

t = - 1.33333 \dots, t = - 0.66666 \dots

Solution steps

t^{2} - 2 t (t + 1) - 8 (t + 1)^{2} = 0

Expand

t^{2} - 2 t (t + 1) - 8 (t + 1)^{2} : - 9 t^{2} - 18 t - 8

- 9 t^{2} - 18 t - 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 ( - 9 ) ( - 8 )}{2 ( - 9 )}

(- 18)^{2} - 4 (- 9) (- 8) = 6

t_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 ( - 9 )}

Separate the solutions

t_{1} = \frac{- ( - 18 ) + 6}{2 ( - 9 )}, t_{2} = \frac{- ( - 18 ) - 6}{2 ( - 9 )}

t = \frac{- ( - 18 ) + 6}{2 ( - 9 )} : - \frac{4}{3}

t = \frac{- ( - 18 ) - 6}{2 ( - 9 )} : - \frac{2}{3}

The solutions to the quadratic equation are:

t = - \frac{4}{3}, t = - \frac{2}{3}

y, y^{2} = \frac{x ^{3}}{2 a - x}

x^{2} = b^{2} + 2 b^{2}

x, x^{2} - 2 x y + y^{2} - 2 x - 2 y = 0

x - x^{2} = 1

n^{2} - n = 12

What are the solutions to the equation t^2-2t(t+1)-8(t+1)^2=0 ?
The solutions to the equation t^2-2t(t+1)-8(t+1)^2=0 are t=-4/3 ,t=-2/3
Find the zeros of t^2-2t(t+1)-8(t+1)^2=0
The zeros of t^2-2t(t+1)-8(t+1)^2=0 are t=-4/3 ,t=-2/3