What are the solutions to the equation 5y^2+16=42y ?

The solutions to the equation 5y^2+16=42y are y=8,y= 2/5

Find the zeros of 5y^2+16=42y

The zeros of 5y^2+16=42y are y=8,y= 2/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 y^{2} + 16 = 42 y

y = 8, y = \frac{2}{5}

Decimal Notation

y = 8, y = 0.4

Solution steps

5 y^{2} + 16 = 42 y

Move

42 y

to the left side

5 y^{2} + 16 - 42 y = 0

Write in the standard form

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

5 y^{2} - 42 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 5 \cdot 16}{2 \cdot 5}

(- 42)^{2} - 4 \cdot 5 \cdot 16 = 38

y_{1, 2} = \frac{- ( - 42 ) \pm 38}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- ( - 42 ) + 38}{2 \cdot 5}, y_{2} = \frac{- ( - 42 ) - 38}{2 \cdot 5}

y = \frac{- ( - 42 ) + 38}{2 \cdot 5} : 8

y = \frac{- ( - 42 ) - 38}{2 \cdot 5} : \frac{2}{5}

The solutions to the quadratic equation are:

y = 8, y = \frac{2}{5}

8 x^{2} = (x - 5)^{2}

(2 - (4))^{2} + (1 - (- 3))^{2} = r^{2}

2 x^{2} + 17 x + 24 = 0

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

4 (6 - 4 x)^{2} = 49

What are the solutions to the equation 5y^2+16=42y ?
The solutions to the equation 5y^2+16=42y are y=8,y= 2/5
Find the zeros of 5y^2+16=42y
The zeros of 5y^2+16=42y are y=8,y= 2/5