What are the solutions to the equation 0=-(8/3)x^2+48x-192 ?

The solutions to the equation 0=-(8/3)x^2+48x-192 are x=6,x=12

Find the zeros of 0=-(8/3)x^2+48x-192

The zeros of 0=-(8/3)x^2+48x-192 are x=6,x=12

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0 = - (\frac{8}{3}) x^{2} + 48 x - 192

x = 6, x = 12

Solution steps

0 = - (\frac{8}{3}) x^{2} + 48 x - 192

Refine

0 = - \frac{8}{3} x^{2} + 48 x - 192

Switch sides

- \frac{8}{3} x^{2} + 48 x - 192 = 0

Write in the standard form

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

- \frac{8 x ^{2}}{3} + 48 x - 192 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - \frac{8}{3} ) ( - 192 )}{2 ( - \frac{8}{3} )}

4 8^{2} - 4 (- \frac{8}{3}) (- 192) = 16

x_{1, 2} = \frac{- 48 \pm 16}{2 ( - \frac{8}{3} )}

Separate the solutions

x_{1} = \frac{- 48 + 16}{2 ( - \frac{8}{3} )}, x_{2} = \frac{- 48 - 16}{2 ( - \frac{8}{3} )}

x = \frac{- 48 + 16}{2 ( - \frac{8}{3} )} : 6

x = \frac{- 48 - 16}{2 ( - \frac{8}{3} )} : 12

The solutions to the quadratic equation are:

x = 6, x = 12

0 = 2 x^{2} - 14 x + 9

15 x^{2} - 8 = 14 x

u^{2} - 12 u

4 x^{2} - 5 = 11 x

x^{2} - 19 = 30

What are the solutions to the equation 0=-(8/3)x^2+48x-192 ?
The solutions to the equation 0=-(8/3)x^2+48x-192 are x=6,x=12
Find the zeros of 0=-(8/3)x^2+48x-192
The zeros of 0=-(8/3)x^2+48x-192 are x=6,x=12