What are the solutions to the equation 3(2x^2-mx)+4nx-2mn=0 ?

The solutions to the equation 3(2x^2-mx)+4nx-2mn=0 are x= m/2 ,x=-(2n)/3

Find the zeros of 3(2x^2-mx)+4nx-2mn=0

The zeros of 3(2x^2-mx)+4nx-2mn=0 are x= m/2 ,x=-(2n)/3

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3 (2 x^{2} - m x) + 4 n x - 2 mn = 0

x = \frac{m}{2}, x = - \frac{2 n}{3}

Solution steps

3 (2 x^{2} - m x) + 4 n x - 2 mn = 0

Expand

3 (2 x^{2} - m x) + 4 n x - 2 mn : 6 x^{2} - 3 m x + 4 n x - 2 mn

6 x^{2} - 3 m x + 4 n x - 2 mn = 0

Write in the standard form

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

6 x^{2} + (- 3 m + 4 n) x - 2 mn = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 m + 4 n ) \pm ( - 3 m + 4 n ) ^{2} - 4 \cdot 6 ( - 2 mn )}{2 \cdot 6}

Simplify

(- 3 m + 4 n)^{2} - 4 \cdot 6 (- 2 mn) : 3 m + 4 n

x_{1, 2} = \frac{- ( - 3 m + 4 n ) \pm ( 3 m + 4 n )}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 3 m + 4 n ) + 3 m + 4 n}{2 \cdot 6}, x_{2} = \frac{- ( - 3 m + 4 n ) - ( 3 m + 4 n )}{2 \cdot 6}

x = \frac{- ( - 3 m + 4 n ) + 3 m + 4 n}{2 \cdot 6} : \frac{m}{2}

x = \frac{- ( - 3 m + 4 n ) - ( 3 m + 4 n )}{2 \cdot 6} : - \frac{2 n}{3}

The solutions to the quadratic equation are:

x = \frac{m}{2}, x = - \frac{2 n}{3}

3 x^{2} - 7 x - 4 = 0

x^{2} - 4 x + 20

6 x^{2} + 2 = 0

x^{2} = 8 x - 35

18 + 3 x - x^{2} = 0

What are the solutions to the equation 3(2x^2-mx)+4nx-2mn=0 ?
The solutions to the equation 3(2x^2-mx)+4nx-2mn=0 are x= m/2 ,x=-(2n)/3
Find the zeros of 3(2x^2-mx)+4nx-2mn=0
The zeros of 3(2x^2-mx)+4nx-2mn=0 are x= m/2 ,x=-(2n)/3