What are the solutions to the equation 4t^3+6t+2.5=25 ?

The solutions to the equation 4t^3+6t+2.5=25 are t= 3/2 ,t=-3/4+i(sqrt(51))/4 ,t=-3/4-i(sqrt(51))/4

Find the zeros of 4t^3+6t+2.5=25

The zeros of 4t^3+6t+2.5=25 are t= 3/2 ,t=-3/4+i(sqrt(51))/4 ,t=-3/4-i(sqrt(51))/4

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4 t^{3} + 6 t + 2.5 = 25

t = \frac{3}{2}, t = - \frac{3}{4} + i \frac{51}{4}, t = - \frac{3}{4} - i \frac{51}{4}

Solution steps

4 t^{3} + 6 t + 2.5 = 25

Multiply both sides by

10

40 t^{3} + 60 t + 25 = 250

Move

250

to the left side

40 t^{3} + 60 t - 225 = 0

Factor

40 t^{3} + 60 t - 225 : 5 (2 t - 3) (4 t^{2} + 6 t + 15)

5 (2 t - 3) (4 t^{2} + 6 t + 15) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 t - 3 = 0 or 4 t^{2} + 6 t + 15 = 0

Solve

2 t - 3 = 0 : t = \frac{3}{2}

Solve

4 t^{2} + 6 t + 15 = 0 : t = - \frac{3}{4} + i \frac{51}{4}, t = - \frac{3}{4} - i \frac{51}{4}

The solutions are

t = \frac{3}{2}, t = - \frac{3}{4} + i \frac{51}{4}, t = - \frac{3}{4} - i \frac{51}{4}

3 x^{2} \cdot 2 x = 162

x (x + 2) (x - 4) = 2

x^{4} + 3 x^{3} - 3 x^{2} - 7 x + 6 = 0

(2 x - 5)^{5} = 243

x^{4} + 16 i = 0

What are the solutions to the equation 4t^3+6t+2.5=25 ?
The solutions to the equation 4t^3+6t+2.5=25 are t= 3/2 ,t=-3/4+i(sqrt(51))/4 ,t=-3/4-i(sqrt(51))/4
Find the zeros of 4t^3+6t+2.5=25
The zeros of 4t^3+6t+2.5=25 are t= 3/2 ,t=-3/4+i(sqrt(51))/4 ,t=-3/4-i(sqrt(51))/4