What are the solutions to the equation 9t^2-48t+54=0 ?

The solutions to the equation 9t^2-48t+54=0 are t=(8+sqrt(10))/3 ,t=(8-sqrt(10))/3

Find the zeros of 9t^2-48t+54=0

The zeros of 9t^2-48t+54=0 are t=(8+sqrt(10))/3 ,t=(8-sqrt(10))/3

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9 t^{2} - 48 t + 54 = 0

t = \frac{8 + 10}{3}, t = \frac{8 - 10}{3}

Decimal Notation

t = 3.72075 \dots, t = 1.61257 \dots

Solution steps

9 t^{2} - 48 t + 54 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 9 \cdot 54}{2 \cdot 9}

(- 48)^{2} - 4 \cdot 9 \cdot 54 = 610

t_{1, 2} = \frac{- ( - 48 ) \pm 6 10}{2 \cdot 9}

Separate the solutions

t_{1} = \frac{- ( - 48 ) + 6 10}{2 \cdot 9}, t_{2} = \frac{- ( - 48 ) - 6 10}{2 \cdot 9}

t = \frac{- ( - 48 ) + 6 10}{2 \cdot 9} : \frac{8 + 10}{3}

t = \frac{- ( - 48 ) - 6 10}{2 \cdot 9} : \frac{8 - 10}{3}

The solutions to the quadratic equation are:

t = \frac{8 + 10}{3}, t = \frac{8 - 10}{3}

24 x^{2} - 36 x = 0

x (x - 2) = (x + 2)

x^{2} - 21 = - 6 x

x^{2} - 5 x + 25 = 10 x

3 (x + 7)^{2} = 48

What are the solutions to the equation 9t^2-48t+54=0 ?
The solutions to the equation 9t^2-48t+54=0 are t=(8+sqrt(10))/3 ,t=(8-sqrt(10))/3
Find the zeros of 9t^2-48t+54=0
The zeros of 9t^2-48t+54=0 are t=(8+sqrt(10))/3 ,t=(8-sqrt(10))/3