What are the solutions to the equation ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1 ?

The solutions to the equation ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1 are t=-4/15 ,t=-1

Find the zeros of ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1

The zeros of ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1 are t=-4/15 ,t=-1

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\frac{( 2 t + 5 ) ( 3 t - 1 )}{3} + \frac{t ^{2} + 5}{2} = \frac{7 t - 5}{6} + 1

t = - \frac{4}{15}, t = - 1

Decimal Notation

t = - 0.26666 \dots, t = - 1

Solution steps

\frac{( 2 t + 5 ) ( 3 t - 1 )}{3} + \frac{t ^{2} + 5}{2} = \frac{7 t - 5}{6} + 1

Find Least Common Multiplier of

3, 2, 6 : 6

Multiply by LCM=

6

\frac{( 2 t + 5 ) ( 3 t - 1 )}{3} \cdot 6 + \frac{t ^{2} + 5}{2} \cdot 6 = \frac{7 t - 5}{6} \cdot 6 + 1 \cdot 6

Simplify

2 (2 t + 5) (3 t - 1) + 3 (t^{2} + 5) = 7 t - 5 + 6

2 (2 t + 5) (3 t - 1) + 3 (t^{2} + 5) = 7 t + 1

Expand

2 (2 t + 5) (3 t - 1) + 3 (t^{2} + 5) : 15 t^{2} + 26 t + 5

15 t^{2} + 26 t + 5 = 7 t + 1

Move

1

to the left side

15 t^{2} + 26 t + 4 = 7 t

Move

7 t

to the left side

15 t^{2} + 19 t + 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 15 \cdot 4}{2 \cdot 15}

1 9^{2} - 4 \cdot 15 \cdot 4 = 11

t_{1, 2} = \frac{- 19 \pm 11}{2 \cdot 15}

Separate the solutions

t_{1} = \frac{- 19 + 11}{2 \cdot 15}, t_{2} = \frac{- 19 - 11}{2 \cdot 15}

t = \frac{- 19 + 11}{2 \cdot 15} : - \frac{4}{15}

t = \frac{- 19 - 11}{2 \cdot 15} : - 1

The solutions to the quadratic equation are:

t = - \frac{4}{15}, t = - 1

∣ 2 x ∣ = x^{2} - 3

4 x^{3} + 12 x^{2} - x - 3

- 1 \cdot \frac{1}{3}

\frac{2}{5} \cdot 6

9 n^{2} - 27 n + 20

What are the solutions to the equation ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1 ?
The solutions to the equation ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1 are t=-4/15 ,t=-1
Find the zeros of ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1
The zeros of ((2t+5)(3t-1))/3+(t^2+5)/2 =(7t-5)/6+1 are t=-4/15 ,t=-1