What are the solutions to the equation 12x^2-55x+28=0 ?

The solutions to the equation 12x^2-55x+28=0 are x=4,x= 7/12

Find the zeros of 12x^2-55x+28=0

The zeros of 12x^2-55x+28=0 are x=4,x= 7/12

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12 x^{2} - 55 x + 28 = 0

x = 4, x = \frac{7}{12}

Decimal Notation

x = 4, x = 0.58333 \dots

Solution steps

12 x^{2} - 55 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 55 ) \pm ( - 55 ) ^{2} - 4 \cdot 12 \cdot 28}{2 \cdot 12}

(- 55)^{2} - 4 \cdot 12 \cdot 28 = 41

x_{1, 2} = \frac{- ( - 55 ) \pm 41}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 55 ) + 41}{2 \cdot 12}, x_{2} = \frac{- ( - 55 ) - 41}{2 \cdot 12}

x = \frac{- ( - 55 ) + 41}{2 \cdot 12} : 4

x = \frac{- ( - 55 ) - 41}{2 \cdot 12} : \frac{7}{12}

The solutions to the quadratic equation are:

x = 4, x = \frac{7}{12}

(2 x - 6)^{2} = 0

0 = 6 x^{2} + x - 1

2 x^{2} - 3 x - 32 = 0

3 x^{2} = 1 + 6 x

x^{2} + 8 x - 15

What are the solutions to the equation 12x^2-55x+28=0 ?
The solutions to the equation 12x^2-55x+28=0 are x=4,x= 7/12
Find the zeros of 12x^2-55x+28=0
The zeros of 12x^2-55x+28=0 are x=4,x= 7/12