What are the solutions to the equation t^2-t-6=4(t-3) ?

The solutions to the equation t^2-t-6=4(t-3) are t=3,t=2

Find the zeros of t^2-t-6=4(t-3)

The zeros of t^2-t-6=4(t-3) are t=3,t=2

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

t = 3, t = 2

Solution steps

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

Expand

4 (t - 3) : 4 t - 12

t^{2} - t - 6 = 4 t - 12

Move

12

to the left side

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

Move

4 t

to the left side

t^{2} - 5 t + 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 6 = 1

t_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 1}

t = \frac{- ( - 5 ) + 1}{2 \cdot 1} : 3

t = \frac{- ( - 5 ) - 1}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

t = 3, t = 2

6 x^{2} - 9 x - 15 = 0

2 x (3 x - 5) = x

(x - 4)^{2} (x + 2) = (x + 2)^{2} (x - 4)

x (x^{2} + 2 x + 3) = x^{3} - 1

2 x^{2} = - 9 x - 10

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