What are the solutions to the equation 2f(a)=3x^2-x+2 ?

The solutions to the equation 2f(a)=3x^2-x+2 are x=(1+sqrt(24f(a)-23))/6 ,x=(1-sqrt(24f(a)-23))/6

Find the zeros of 2f(a)=3x^2-x+2

The zeros of 2f(a)=3x^2-x+2 are x=(1+sqrt(24f(a)-23))/6 ,x=(1-sqrt(24f(a)-23))/6

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2 f (a) = 3 x^{2} - x + 2

x = \frac{1 + 24 f ( a ) - 23}{6}, x = \frac{1 - 24 f ( a ) - 23}{6}

Solution steps

2 f (a) = 3 x^{2} - x + 2

Switch sides

3 x^{2} - x + 2 = 2 f (a)

Move

2 f (a)

to the left side

3 x^{2} - x + 2 - 2 f (a) = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 3 (2 - 2 f (a)) : 24 f (a) - 23

x_{1, 2} = \frac{- ( - 1 ) \pm 24 f ( a ) - 23}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 24 f ( a ) - 23}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 24 f ( a ) - 23}{2 \cdot 3}

x = \frac{- ( - 1 ) + 24 f ( a ) - 23}{2 \cdot 3} : \frac{1 + 24 f ( a ) - 23}{6}

x = \frac{- ( - 1 ) - 24 f ( a ) - 23}{2 \cdot 3} : \frac{1 - 24 f ( a ) - 23}{6}

The solutions to the quadratic equation are:

x = \frac{1 + 24 f ( a ) - 23}{6}, x = \frac{1 - 24 f ( a ) - 23}{6}

12 x + 3 = - 4 x^{2} + 20

(x + 3) 2 = 8 (x - 2) + x^{2}

x^{2} + 13 x + 24 = 0

2 b^{2} + 40 = 0

(\frac{5}{13})^{2} + x^{2} = 1

What are the solutions to the equation 2f(a)=3x^2-x+2 ?
The solutions to the equation 2f(a)=3x^2-x+2 are x=(1+sqrt(24f(a)-23))/6 ,x=(1-sqrt(24f(a)-23))/6
Find the zeros of 2f(a)=3x^2-x+2
The zeros of 2f(a)=3x^2-x+2 are x=(1+sqrt(24f(a)-23))/6 ,x=(1-sqrt(24f(a)-23))/6