What are the solutions to the equation d(x)= 1/2 x^2-10x ?

The solutions to the equation d(x)= 1/2 x^2-10x are x=0,x=20+2d

Find the zeros of d(x)= 1/2 x^2-10x

The zeros of d(x)= 1/2 x^2-10x are x=0,x=20+2d

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

d (x) = \frac{1}{2} x^{2} - 10 x

x = 0, x = 20 + 2 d

Solution steps

d (x) = \frac{1}{2} x^{2} - 10 x

Multiply both sides by

2

2 d x = x^{2} - 20 x

Switch sides

x^{2} - 20 x = 2 d x

Move

2 d x

to the left side

x^{2} - 20 x - 2 d x = 0

Write in the standard form

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

x^{2} - (20 + 2 d) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 - 2 d ) \pm ( - 20 - 2 d ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

Simplify

(- 20 - 2 d)^{2} - 4 \cdot 1 \cdot 0 : - 2 d - 20

x_{1, 2} = \frac{- ( - 20 - 2 d ) \pm ( - 2 d - 20 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 20 - 2 d ) - 2 d - 20}{2 \cdot 1}, x_{2} = \frac{- ( - 20 - 2 d ) - ( - 2 d - 20 )}{2 \cdot 1}

x = \frac{- ( - 20 - 2 d ) - 2 d - 20}{2 \cdot 1} : 0

x = \frac{- ( - 20 - 2 d ) - ( - 2 d - 20 )}{2 \cdot 1} : 20 + 2 d

The solutions to the quadratic equation are:

x = 0, x = 20 + 2 d

y^{^{''}} - y = 1.7 sin (t)

y^{^{''}} + y^{^{'}} - 12 y = 2 e^{5 t}

\frac{1}{2} x^{^{''}} + 2 x^{^{'}} + 4 x = 20 cos (t)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 40 cos (8 x)

y^{^{''''}} - 4 y^{^{''}} = x^{2} + 3 e^{x}

What are the solutions to the equation d(x)= 1/2 x^2-10x ?
The solutions to the equation d(x)= 1/2 x^2-10x are x=0,x=20+2d
Find the zeros of d(x)= 1/2 x^2-10x
The zeros of d(x)= 1/2 x^2-10x are x=0,x=20+2d