What are the solutions to the equation 3x^2-16x+24=0 ?

The solutions to the equation 3x^2-16x+24=0 are x= 8/3+i(2sqrt(2))/3 ,x= 8/3-i(2sqrt(2))/3

Find the zeros of 3x^2-16x+24=0

The zeros of 3x^2-16x+24=0 are x= 8/3+i(2sqrt(2))/3 ,x= 8/3-i(2sqrt(2))/3

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3 x^{2} - 16 x + 24 = 0

x = \frac{8}{3} + i \frac{2 2}{3}, x = \frac{8}{3} - i \frac{2 2}{3}

Solution steps

3 x^{2} - 16 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 24}{2 \cdot 3}

Simplify

(- 16)^{2} - 4 \cdot 3 \cdot 24 : 42 i

x_{1, 2} = \frac{- ( - 16 ) \pm 4 2 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4 2 i}{2 \cdot 3}, x_{2} = \frac{- ( - 16 ) - 4 2 i}{2 \cdot 3}

x = \frac{- ( - 16 ) + 4 2 i}{2 \cdot 3} : \frac{8}{3} + i \frac{2 2}{3}

x = \frac{- ( - 16 ) - 4 2 i}{2 \cdot 3} : \frac{8}{3} - i \frac{2 2}{3}

The solutions to the quadratic equation are:

x = \frac{8}{3} + i \frac{2 2}{3}, x = \frac{8}{3} - i \frac{2 2}{3}

p, p^{1} \frac{v ^{1}}{t ^{1}} = p^{2} \frac{v ^{2}}{t ^{2}}

(x + 1) - x (x - 3) = 6

x^{2} + 20 x + 99 = 0

3 x^{2} = 8 x

x^{2} - 2 x - 5 = 0

What are the solutions to the equation 3x^2-16x+24=0 ?
The solutions to the equation 3x^2-16x+24=0 are x= 8/3+i(2sqrt(2))/3 ,x= 8/3-i(2sqrt(2))/3
Find the zeros of 3x^2-16x+24=0
The zeros of 3x^2-16x+24=0 are x= 8/3+i(2sqrt(2))/3 ,x= 8/3-i(2sqrt(2))/3