What are the solutions to the equation A(x)=-4/3 x^2+42x ?

The solutions to the equation A(x)=-4/3 x^2+42x are x=0,x=(126-3A)/4

Find the zeros of A(x)=-4/3 x^2+42x

The zeros of A(x)=-4/3 x^2+42x are x=0,x=(126-3A)/4

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A (x) = - \frac{4}{3} x^{2} + 42 x

x = 0, x = \frac{126 - 3 A}{4}

Solution steps

A (x) = - \frac{4}{3} x^{2} + 42 x

Multiply both sides by

3

3 A x = - 4 x^{2} + 126 x

Switch sides

- 4 x^{2} + 126 x = 3 A x

Move

3 A x

to the left side

- 4 x^{2} + 126 x - 3 A x = 0

Write in the standard form

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

- 4 x^{2} + (126 - 3 A) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 126 - 3 A ) \pm ( 126 - 3 A ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

Simplify

(126 - 3 A)^{2} - 4 (- 4) \cdot 0 : - 3 A + 126

x_{1, 2} = \frac{- ( 126 - 3 A ) \pm ( - 3 A + 126 )}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( 126 - 3 A ) - 3 A + 126}{2 ( - 4 )}, x_{2} = \frac{- ( 126 - 3 A ) - ( - 3 A + 126 )}{2 ( - 4 )}

x = \frac{- ( 126 - 3 A ) - 3 A + 126}{2 ( - 4 )} : 0

x = \frac{- ( 126 - 3 A ) - ( - 3 A + 126 )}{2 ( - 4 )} : \frac{126 - 3 A}{4}

The solutions to the quadratic equation are:

x = 0, x = \frac{126 - 3 A}{4}

12 x^{2} - 24 x + 6 = 0

6 x^{2} + 7 x - 15 = 0

- 12 x^{2} + 18 x - 6 = 0

2 = 3.2 + 2.8 x - 0.04 x^{2}

(x^{2} + 2 x - 1) = 0

What are the solutions to the equation A(x)=-4/3 x^2+42x ?
The solutions to the equation A(x)=-4/3 x^2+42x are x=0,x=(126-3A)/4
Find the zeros of A(x)=-4/3 x^2+42x
The zeros of A(x)=-4/3 x^2+42x are x=0,x=(126-3A)/4