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

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

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

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

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

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

Decimal Notation

x = 2, x = 0.66666 \dots

Solution steps

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

Switch sides

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}

x^{2} - 9 = - 5

x^{4} y^{3} - x^{5} y^{4}

c (c - 6) = 16

\frac{2}{8} + \frac{5}{12}

x^{2} + x = 1

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