What are the solutions to the equation 36x^2-48x+96=0 ?

The solutions to the equation 36x^2-48x+96=0 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3

Find the zeros of 36x^2-48x+96=0

The zeros of 36x^2-48x+96=0 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3

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36 x^{2} - 48 x + 96 = 0

x = \frac{2}{3} + i \frac{2 5}{3}, x = \frac{2}{3} - i \frac{2 5}{3}

Solution steps

36 x^{2} - 48 x + 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 36 \cdot 96}{2 \cdot 36}

Simplify

(- 48)^{2} - 4 \cdot 36 \cdot 96 : 485 i

x_{1, 2} = \frac{- ( - 48 ) \pm 48 5 i}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 48 5 i}{2 \cdot 36}, x_{2} = \frac{- ( - 48 ) - 48 5 i}{2 \cdot 36}

x = \frac{- ( - 48 ) + 48 5 i}{2 \cdot 36} : \frac{2}{3} + i \frac{2 5}{3}

x = \frac{- ( - 48 ) - 48 5 i}{2 \cdot 36} : \frac{2}{3} - i \frac{2 5}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{2 5}{3}, x = \frac{2}{3} - i \frac{2 5}{3}

- \frac{1}{6} - (- \frac{8}{9})

r^{2} + 4 r + 29 = 0

2 (2 x + 3) = x - 13

x^{2} - 17 x + 42

4 r^{2} + r - 3

What are the solutions to the equation 36x^2-48x+96=0 ?
The solutions to the equation 36x^2-48x+96=0 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3
Find the zeros of 36x^2-48x+96=0
The zeros of 36x^2-48x+96=0 are x= 2/3+i(2sqrt(5))/3 ,x= 2/3-i(2sqrt(5))/3