What are the solutions to the equation 2f(a)=4x^2-x+5 ?

The solutions to the equation 2f(a)=4x^2-x+5 are x=(1+sqrt(32f(a)-79))/8 ,x=(1-sqrt(32f(a)-79))/8

Find the zeros of 2f(a)=4x^2-x+5

The zeros of 2f(a)=4x^2-x+5 are x=(1+sqrt(32f(a)-79))/8 ,x=(1-sqrt(32f(a)-79))/8

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2 f (a) = 4 x^{2} - x + 5

x = \frac{1 + 32 f ( a ) - 79}{8}, x = \frac{1 - 32 f ( a ) - 79}{8}

Solution steps

2 f (a) = 4 x^{2} - x + 5

Switch sides

4 x^{2} - x + 5 = 2 f (a)

Move

2 f (a)

to the left side

4 x^{2} - x + 5 - 2 f (a) = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 4 ( 5 - 2 f ( a ) )}{2 \cdot 4}

Simplify

(- 1)^{2} - 4 \cdot 4 (5 - 2 f (a)) : 32 f (a) - 79

x_{1, 2} = \frac{- ( - 1 ) \pm 32 f ( a ) - 79}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 32 f ( a ) - 79}{2 \cdot 4}, x_{2} = \frac{- ( - 1 ) - 32 f ( a ) - 79}{2 \cdot 4}

x = \frac{- ( - 1 ) + 32 f ( a ) - 79}{2 \cdot 4} : \frac{1 + 32 f ( a ) - 79}{8}

x = \frac{- ( - 1 ) - 32 f ( a ) - 79}{2 \cdot 4} : \frac{1 - 32 f ( a ) - 79}{8}

The solutions to the quadratic equation are:

x = \frac{1 + 32 f ( a ) - 79}{8}, x = \frac{1 - 32 f ( a ) - 79}{8}

x^{2} - 18 x + 7 = 0

2 x^{2} - 28 x = - 76

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

x^{2} - 6 x + 6 = 0

9 t^{2} - 45 t + 54 = 0

What are the solutions to the equation 2f(a)=4x^2-x+5 ?
The solutions to the equation 2f(a)=4x^2-x+5 are x=(1+sqrt(32f(a)-79))/8 ,x=(1-sqrt(32f(a)-79))/8
Find the zeros of 2f(a)=4x^2-x+5
The zeros of 2f(a)=4x^2-x+5 are x=(1+sqrt(32f(a)-79))/8 ,x=(1-sqrt(32f(a)-79))/8