What are the solutions to the equation 4.9x^2-42x+80=0 ?

The solutions to the equation 4.9x^2-42x+80=0 are x= 40/7 ,x= 20/7

Find the zeros of 4.9x^2-42x+80=0

The zeros of 4.9x^2-42x+80=0 are x= 40/7 ,x= 20/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4.9 x^{2} - 42 x + 80 = 0

x = \frac{40}{7}, x = \frac{20}{7}

Decimal Notation

x = 5.71428 \dots, x = 2.85714 \dots

Solution steps

4.9 x^{2} - 42 x + 80 = 0

Multiply both sides by

10

49 x^{2} - 420 x + 800 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 420 ) \pm ( - 420 ) ^{2} - 4 \cdot 49 \cdot 800}{2 \cdot 49}

(- 420)^{2} - 4 \cdot 49 \cdot 800 = 140

x_{1, 2} = \frac{- ( - 420 ) \pm 140}{2 \cdot 49}

Separate the solutions

x_{1} = \frac{- ( - 420 ) + 140}{2 \cdot 49}, x_{2} = \frac{- ( - 420 ) - 140}{2 \cdot 49}

x = \frac{- ( - 420 ) + 140}{2 \cdot 49} : \frac{40}{7}

x = \frac{- ( - 420 ) - 140}{2 \cdot 49} : \frac{20}{7}

The solutions to the quadratic equation are:

x = \frac{40}{7}, x = \frac{20}{7}

a x^{2} + 9 a x + 20 = 0

4 x = - 5 x^{2} - 1

90 = - 3.7829 x^{2} + 35.044 x - 1.4862

x^{2} + (- 4) x + (- 4) + 8 = 0

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

What are the solutions to the equation 4.9x^2-42x+80=0 ?
The solutions to the equation 4.9x^2-42x+80=0 are x= 40/7 ,x= 20/7
Find the zeros of 4.9x^2-42x+80=0
The zeros of 4.9x^2-42x+80=0 are x= 40/7 ,x= 20/7