What are the solutions to the equation 3t(2t-1)=4t+3 ?

The solutions to the equation 3t(2t-1)=4t+3 are t= 3/2 ,t=-1/3

Find the zeros of 3t(2t-1)=4t+3

The zeros of 3t(2t-1)=4t+3 are t= 3/2 ,t=-1/3

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3 t (2 t - 1) = 4 t + 3

t = \frac{3}{2}, t = - \frac{1}{3}

Decimal

t = 1.5, t = - 0.33333 \dots

Solution steps

3 t (2 t - 1) = 4 t + 3

Expand

3 t (2 t - 1) : 6 t^{2} - 3 t

6 t^{2} - 3 t = 4 t + 3

Move

3

to the left side

6 t^{2} - 3 t - 3 = 4 t

Move

4 t

to the left side

6 t^{2} - 7 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 (- 3) = 11

t_{1, 2} = \frac{- ( - 7 ) \pm 11}{2 \cdot 6}

Separate the solutions

t_{1} = \frac{- ( - 7 ) + 11}{2 \cdot 6}, t_{2} = \frac{- ( - 7 ) - 11}{2 \cdot 6}

t = \frac{- ( - 7 ) + 11}{2 \cdot 6} : \frac{3}{2}

t = \frac{- ( - 7 ) - 11}{2 \cdot 6} : - \frac{1}{3}

The solutions to the quadratic equation are:

t = \frac{3}{2}, t = - \frac{1}{3}

v^{2} = - 13 v

2 X^{2} + 13 X = 7

x^{2} - 5 x + 6 = 4

x^{2} - 5 x + 6 = 9

2 r^{2} - 72 = 0

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