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

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

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

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

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

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

Solution steps

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

Multiply both sides by

10

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

Switch sides

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}

x^{3} - 3 a + 2 = 0

4 x^{3} - \frac{4 x}{9} = 0

q^{4} - 2 = 1

2 z (z + 3) (z + 5) = 0

32 x^{5} - 40 x^{3} + 10 x + 1 = 0

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