What are the solutions to the equation 9t(t+2)-3t(t-2)=2(t+4)(t+6) ?

The solutions to the equation 9t(t+2)-3t(t-2)=2(t+4)(t+6) are t=-4,t=3

Find the zeros of 9t(t+2)-3t(t-2)=2(t+4)(t+6)

The zeros of 9t(t+2)-3t(t-2)=2(t+4)(t+6) are t=-4,t=3

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9 t (t + 2) - 3 t (t - 2) = 2 (t + 4) (t + 6)

t = - 4, t = 3

Solution steps

9 t (t + 2) - 3 t (t - 2) = 2 (t + 4) (t + 6)

Expand

9 t (t + 2) - 3 t (t - 2) : 6 t^{2} + 24 t

Expand

2 (t + 4) (t + 6) : 2 t^{2} + 20 t + 48

6 t^{2} + 24 t = 2 t^{2} + 20 t + 48

Switch sides

2 t^{2} + 20 t + 48 = 6 t^{2} + 24 t

Move

24 t

to the left side

2 t^{2} - 4 t + 48 = 6 t^{2}

Move

6 t^{2}

to the left side

- 4 t^{2} - 4 t + 48 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 4 ) \cdot 48}{2 ( - 4 )}

(- 4)^{2} - 4 (- 4) \cdot 48 = 28

t_{1, 2} = \frac{- ( - 4 ) \pm 28}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 28}{2 ( - 4 )}, t_{2} = \frac{- ( - 4 ) - 28}{2 ( - 4 )}

t = \frac{- ( - 4 ) + 28}{2 ( - 4 )} : - 4

t = \frac{- ( - 4 ) - 28}{2 ( - 4 )} : 3

The solutions to the quadratic equation are:

t = - 4, t = 3

\frac{12 !}{( 12 - 4 ) ! 4 !}

x^{3} - 3 x^{2}

\frac{3 ^{- 2} a ^{0}}{b ^{- 2}}

x + 4 \geq \frac{1}{2} (x - 3)

\frac{n}{n + 1} - 1

What are the solutions to the equation 9t(t+2)-3t(t-2)=2(t+4)(t+6) ?
The solutions to the equation 9t(t+2)-3t(t-2)=2(t+4)(t+6) are t=-4,t=3
Find the zeros of 9t(t+2)-3t(t-2)=2(t+4)(t+6)
The zeros of 9t(t+2)-3t(t-2)=2(t+4)(t+6) are t=-4,t=3