What are the solutions to the equation 160-104x+12x^2=0 ?

The solutions to the equation 160-104x+12x^2=0 are x= 20/3 ,x=2

Find the zeros of 160-104x+12x^2=0

The zeros of 160-104x+12x^2=0 are x= 20/3 ,x=2

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160 - 104 x + 12 x^{2} = 0

x = \frac{20}{3}, x = 2

Decimal Notation

x = 6.66666 \dots, x = 2

Solution steps

160 - 104 x + 12 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

12 x^{2} - 104 x + 160 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 104 ) \pm ( - 104 ) ^{2} - 4 \cdot 12 \cdot 160}{2 \cdot 12}

(- 104)^{2} - 4 \cdot 12 \cdot 160 = 56

x_{1, 2} = \frac{- ( - 104 ) \pm 56}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 104 ) + 56}{2 \cdot 12}, x_{2} = \frac{- ( - 104 ) - 56}{2 \cdot 12}

x = \frac{- ( - 104 ) + 56}{2 \cdot 12} : \frac{20}{3}

x = \frac{- ( - 104 ) - 56}{2 \cdot 12} : 2

The solutions to the quadratic equation are:

x = \frac{20}{3}, x = 2

\frac{1}{2} x^{2} + \frac{1}{4} x - \frac{1}{6} = 0

n^{2} - 16 = 0

x^{2} + 10 x + 25

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

x^{2} + 1.2 x + 0.56 = 0

What are the solutions to the equation 160-104x+12x^2=0 ?
The solutions to the equation 160-104x+12x^2=0 are x= 20/3 ,x=2
Find the zeros of 160-104x+12x^2=0
The zeros of 160-104x+12x^2=0 are x= 20/3 ,x=2