What are the solutions to the equation 192-112x+12x^2=0 ?

The solutions to the equation 192-112x+12x^2=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3

Find the zeros of 192-112x+12x^2=0

The zeros of 192-112x+12x^2=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3

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192 - 112 x + 12 x^{2} = 0

x = \frac{2 ( 7 + 13 )}{3}, x = \frac{2 ( 7 - 13 )}{3}

Decimal Notation

x = 7.07036 \dots, x = 2.26296 \dots

Solution steps

192 - 112 x + 12 x^{2} = 0

Write in the standard form

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

12 x^{2} - 112 x + 192 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 112 ) \pm ( - 112 ) ^{2} - 4 \cdot 12 \cdot 192}{2 \cdot 12}

(- 112)^{2} - 4 \cdot 12 \cdot 192 = 1613

x_{1, 2} = \frac{- ( - 112 ) \pm 16 13}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 112 ) + 16 13}{2 \cdot 12}, x_{2} = \frac{- ( - 112 ) - 16 13}{2 \cdot 12}

x = \frac{- ( - 112 ) + 16 13}{2 \cdot 12} : \frac{2 ( 7 + 13 )}{3}

x = \frac{- ( - 112 ) - 16 13}{2 \cdot 12} : \frac{2 ( 7 - 13 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 7 + 13 )}{3}, x = \frac{2 ( 7 - 13 )}{3}

64 - \frac{x ^{2}}{25} = 0

x^{2} - 16 x + 55 = 0

8 y^{2} = 16 y

45 x^{2} = 0

v^{2} - 10 v + 24 = 0

What are the solutions to the equation 192-112x+12x^2=0 ?
The solutions to the equation 192-112x+12x^2=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3
Find the zeros of 192-112x+12x^2=0
The zeros of 192-112x+12x^2=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3