What are the solutions to the equation 36x^2-96x+48=0 ?

The solutions to the equation 36x^2-96x+48=0 are x=2,x= 2/3

Find the zeros of 36x^2-96x+48=0

The zeros of 36x^2-96x+48=0 are x=2,x= 2/3

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36 x^{2} - 96 x + 48 = 0

x = 2, x = \frac{2}{3}

Solution steps

36 x^{2} - 96 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 96 ) \pm ( - 96 ) ^{2} - 4 \cdot 36 \cdot 48}{2 \cdot 36}

(- 96)^{2} - 4 \cdot 36 \cdot 48 = 48

x_{1, 2} = \frac{- ( - 96 ) \pm 48}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- ( - 96 ) + 48}{2 \cdot 36}, x_{2} = \frac{- ( - 96 ) - 48}{2 \cdot 36}

x = \frac{- ( - 96 ) + 48}{2 \cdot 36} : 2

x = \frac{- ( - 96 ) - 48}{2 \cdot 36} : \frac{2}{3}

The solutions to the quadratic equation are:

x = 2, x = \frac{2}{3}

(3 x - 10)^{2} = 0

\frac{\frac{π}{6} + \frac{π}{3}}{2}

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

2 (x - 3) = \frac{1}{2} (4 x - 12)

What are the solutions to the equation 36x^2-96x+48=0 ?
The solutions to the equation 36x^2-96x+48=0 are x=2,x= 2/3
Find the zeros of 36x^2-96x+48=0
The zeros of 36x^2-96x+48=0 are x=2,x= 2/3