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

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

Find the zeros of (3t+5)(t-2)=-6

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

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(3 t + 5) (t - 2) = - 6

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

Decimal Notation

t = 1.33333 \dots, t = - 1

Solution steps

(3 t + 5) (t - 2) = - 6

Expand

(3 t + 5) (t - 2) : 3 t^{2} - t - 10

3 t^{2} - t - 10 = - 6

Move

6

to the left side

3 t^{2} - t - 4 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 3 (- 4) = 7

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

Separate the solutions

t_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 3}, t_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 3}

t = \frac{- ( - 1 ) + 7}{2 \cdot 3} : \frac{4}{3}

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

The solutions to the quadratic equation are:

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

x^{2} + 3 x + 20 = 0

5 x^{2} + 21 x + 9 = 5, 5 x^{2} + 21 x + 9 = 5

- 6 x^{2} + 24 x - 30 = 0

- 4.9 x^{2} + 26 x = 0

8^{2} + x^{2} = 6^{2}

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