What are the solutions to the equation 6t(t+4)=t-20 ?

The solutions to the equation 6t(t+4)=t-20 are t=-4/3 ,t=-5/2

Find the zeros of 6t(t+4)=t-20

The zeros of 6t(t+4)=t-20 are t=-4/3 ,t=-5/2

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6 t (t + 4) = t - 20

t = - \frac{4}{3}, t = - \frac{5}{2}

Decimal Notation

t = - 1.33333 \dots, t = - 2.5

Solution steps

6 t (t + 4) = t - 20

Expand

6 t (t + 4) : 6 t^{2} + 24 t

6 t^{2} + 24 t = t - 20

Move

20

to the left side

6 t^{2} + 24 t + 20 = t

Move

t

to the left side

6 t^{2} + 23 t + 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 \cdot 6 \cdot 20}{2 \cdot 6}

2 3^{2} - 4 \cdot 6 \cdot 20 = 7

t_{1, 2} = \frac{- 23 \pm 7}{2 \cdot 6}

Separate the solutions

t_{1} = \frac{- 23 + 7}{2 \cdot 6}, t_{2} = \frac{- 23 - 7}{2 \cdot 6}

t = \frac{- 23 + 7}{2 \cdot 6} : - \frac{4}{3}

t = \frac{- 23 - 7}{2 \cdot 6} : - \frac{5}{2}

The solutions to the quadratic equation are:

t = - \frac{4}{3}, t = - \frac{5}{2}

25 6^{\frac{5}{4}}

(55)^{2}

100 = 48.0433 1^{2} + 2 \cdot x \cdot (80 - 32)

- 5 x = 9 - 2 x

a^{3} b^{3} - 64

What are the solutions to the equation 6t(t+4)=t-20 ?
The solutions to the equation 6t(t+4)=t-20 are t=-4/3 ,t=-5/2
Find the zeros of 6t(t+4)=t-20
The zeros of 6t(t+4)=t-20 are t=-4/3 ,t=-5/2