What are the solutions to the equation G(x)=-3x^2+102x ?

The solutions to the equation G(x)=-3x^2+102x are x=0,x=(102-G)/3

Find the zeros of G(x)=-3x^2+102x

The zeros of G(x)=-3x^2+102x are x=0,x=(102-G)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

G (x) = - 3 x^{2} + 102 x

x = 0, x = \frac{102 - G}{3}

Solution steps

G (x) = - 3 x^{2} + 102 x

Refine

G x = - 3 x^{2} + 102 x

Switch sides

- 3 x^{2} + 102 x = G x

Move

G x

to the left side

- 3 x^{2} + 102 x - G x = 0

Write in the standard form

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

- 3 x^{2} + (102 - G) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 102 - G ) \pm ( 102 - G ) ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

Simplify

(102 - G)^{2} - 4 (- 3) \cdot 0 : - G + 102

x_{1, 2} = \frac{- ( 102 - G ) \pm ( - G + 102 )}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( 102 - G ) - G + 102}{2 ( - 3 )}, x_{2} = \frac{- ( 102 - G ) - ( - G + 102 )}{2 ( - 3 )}

x = \frac{- ( 102 - G ) - G + 102}{2 ( - 3 )} : 0

x = \frac{- ( 102 - G ) - ( - G + 102 )}{2 ( - 3 )} : \frac{102 - G}{3}

The solutions to the quadratic equation are:

x = 0, x = \frac{102 - G}{3}

y - 5 = \frac{1}{4} (y - 3)^{2}

x^{2} + x + 8 = 5

9 x^{2} - 15 x + 8 = 0

x^{2} - \frac{1}{2} x + \frac{1}{4} = 0

14 x^{2} = 80 x

What are the solutions to the equation G(x)=-3x^2+102x ?
The solutions to the equation G(x)=-3x^2+102x are x=0,x=(102-G)/3
Find the zeros of G(x)=-3x^2+102x
The zeros of G(x)=-3x^2+102x are x=0,x=(102-G)/3