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

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

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

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

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

t = - 1, t = \frac{4}{3}

Decimal

t = - 1, t = 1.33333 \dots

Solution steps

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

Move

4 (t + 1)

to the left side

3 t (t + 1) - 4 (t + 1) = 0

Factor

3 t (t + 1) - 4 (t + 1) : (t + 1) (3 t - 4)

(t + 1) (3 t - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

t + 1 = 0 or 3 t - 4 = 0

Solve

t + 1 = 0 : t = - 1

Solve

3 t - 4 = 0 : t = \frac{4}{3}

The solutions to the quadratic equation are:

t = - 1, t = \frac{4}{3}

x^{2} + 22 x + 20 = 3 x^{2}

x^{2} + 12 x + 63 = 0

x^{2} - 2 x - 4 = - 2 x

x^{2} - 12 x = 0

3 x - 2 = x^{2} + 5 x - 2

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