What are the solutions to the equation 3(t^2-9)^2+16(t^2-9)=-5 ?

The solutions to the equation 3(t^2-9)^2+16(t^2-9)=-5 are t=sqrt(26/3),t=-sqrt(26/3),t=2,t=-2

Find the zeros of 3(t^2-9)^2+16(t^2-9)=-5

The zeros of 3(t^2-9)^2+16(t^2-9)=-5 are t=sqrt(26/3),t=-sqrt(26/3),t=2,t=-2

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3 (t^{2} - 9)^{2} + 16 (t^{2} - 9) = - 5

t = \frac{26}{3}, t = - \frac{26}{3}, t = 2, t = - 2

Decimal Notation

t = 2.94392 \dots, t = - 2.94392 \dots, t = 2, t = - 2

Solution steps

3 (t^{2} - 9)^{2} + 16 (t^{2} - 9) = - 5

Expand

3 (t^{2} - 9)^{2} + 16 (t^{2} - 9) : 3 t^{4} - 38 t^{2} + 99

3 t^{4} - 38 t^{2} + 99 = - 5

Move

5

to the left side

3 t^{4} - 38 t^{2} + 104 = 0

Rewrite the equation with

u = t^{2}

and

u^{2} = t^{4}

3 u^{2} - 38 u + 104 = 0

Solve

3 u^{2} - 38 u + 104 = 0 : u = \frac{26}{3}, u = 4

u = \frac{26}{3}, u = 4

Substitute back

u = t^{2},

solve for

t

Solve

t^{2} = \frac{26}{3} : t = \frac{26}{3}, t = - \frac{26}{3}

Solve

t^{2} = 4 : t = 2, t = - 2

The solutions are

t = \frac{26}{3}, t = - \frac{26}{3}, t = 2, t = - 2

D, 2 b - hb = D^{3}

x^{3} - 6 x^{2} + x - 6 = x^{2} + 1

y (y - 5)^{3} - 14 (y - 5)^{2} = 0

b^{3} = 4

- 7 x - (x + 6) = 9 + 2 x^{9} + 2 x

What are the solutions to the equation 3(t^2-9)^2+16(t^2-9)=-5 ?
The solutions to the equation 3(t^2-9)^2+16(t^2-9)=-5 are t=sqrt(26/3),t=-sqrt(26/3),t=2,t=-2
Find the zeros of 3(t^2-9)^2+16(t^2-9)=-5
The zeros of 3(t^2-9)^2+16(t^2-9)=-5 are t=sqrt(26/3),t=-sqrt(26/3),t=2,t=-2