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

8. 5^{2}

- 5 (x - 2) - (x + 2) = 50

\frac{1}{3} x + \frac{5}{6} = 2 x

2 - 3 (2 (2 x + 1) - 3 x) = 5 - (2 x + 1)

15 a^{2} b + 2 ab - 6 b a^{2} + 8 b

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