What are the solutions to the equation 36x^2-48x+12=0 ?

The solutions to the equation 36x^2-48x+12=0 are x=1,x= 1/3

Find the zeros of 36x^2-48x+12=0

The zeros of 36x^2-48x+12=0 are x=1,x= 1/3

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36 x^{2} - 48 x + 12 = 0

x = 1, x = \frac{1}{3}

Solution steps

36 x^{2} - 48 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 36 \cdot 12}{2 \cdot 36}

(- 48)^{2} - 4 \cdot 36 \cdot 12 = 24

x_{1, 2} = \frac{- ( - 48 ) \pm 24}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 24}{2 \cdot 36}, x_{2} = \frac{- ( - 48 ) - 24}{2 \cdot 36}

x = \frac{- ( - 48 ) + 24}{2 \cdot 36} : 1

x = \frac{- ( - 48 ) - 24}{2 \cdot 36} : \frac{1}{3}

The solutions to the quadratic equation are:

x = 1, x = \frac{1}{3}

\frac{1 + x}{2} < \frac{x - 1}{3}

2 x + 4 > 0

8 x - 7 = 15 + 19 x

2 x + 8 = 2 (x + 4)

1 1^{2}

What are the solutions to the equation 36x^2-48x+12=0 ?
The solutions to the equation 36x^2-48x+12=0 are x=1,x= 1/3
Find the zeros of 36x^2-48x+12=0
The zeros of 36x^2-48x+12=0 are x=1,x= 1/3