What are the solutions to the equation solvefor x,12x^2-15xy=0 ?

The solutions to the equation solvefor x,12x^2-15xy=0 are x=(5y)/4 ,x=0

Find the zeros of solvefor x,12x^2-15xy=0

The zeros of solvefor x,12x^2-15xy=0 are x=(5y)/4 ,x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, 12 x^{2} - 15 x y = 0

x = \frac{5 y}{4}, x = 0

Solution steps

12 x^{2} - 15 x y = 0

Write in the standard form

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

12 x^{2} - 15 y x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 y ) \pm ( - 15 y ) ^{2} - 4 \cdot 12 \cdot 0}{2 \cdot 12}

Simplify

(- 15 y)^{2} - 4 \cdot 12 \cdot 0 : 15 y

x_{1, 2} = \frac{- ( - 15 y ) \pm 15 y}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 15 y ) + 15 y}{2 \cdot 12}, x_{2} = \frac{- ( - 15 y ) - 15 y}{2 \cdot 12}

x = \frac{- ( - 15 y ) + 15 y}{2 \cdot 12} : \frac{5 y}{4}

x = \frac{- ( - 15 y ) - 15 y}{2 \cdot 12} : 0

The solutions to the quadratic equation are:

x = \frac{5 y}{4}, x = 0

x, 4 x^{2} + 2 y^{2} = 4 x y + 2 x y - 1

15 x^{2} - 28 = x

x, - 8 x^{2} + 18 y^{2} + 18 z^{2} = 2

t, h = \frac{1}{2} \cdot a_{g (t)} \cdot t^{2}

(7 x)^{2} - 49 = 0

What are the solutions to the equation solvefor x,12x^2-15xy=0 ?
The solutions to the equation solvefor x,12x^2-15xy=0 are x=(5y)/4 ,x=0
Find the zeros of solvefor x,12x^2-15xy=0
The zeros of solvefor x,12x^2-15xy=0 are x=(5y)/4 ,x=0