What are the solutions to the equation 6c^2+4c-13=5c^2 ?

The solutions to the equation 6c^2+4c-13=5c^2 are c=-2+sqrt(17),c=-2-sqrt(17)

Find the zeros of 6c^2+4c-13=5c^2

The zeros of 6c^2+4c-13=5c^2 are c=-2+sqrt(17),c=-2-sqrt(17)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 c^{2} + 4 c - 13 = 5 c^{2}

c = - 2 + 17, c = - 2 - 17

Decimal Notation

c = 2.12310 \dots, c = - 6.12310 \dots

Solution steps

6 c^{2} + 4 c - 13 = 5 c^{2}

Move

5 c^{2}

to the left side

c^{2} + 4 c - 13 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 13) = 217

c_{1, 2} = \frac{- 4 \pm 2 17}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 4 + 2 17}{2 \cdot 1}, c_{2} = \frac{- 4 - 2 17}{2 \cdot 1}

c = \frac{- 4 + 2 17}{2 \cdot 1} : - 2 + 17

c = \frac{- 4 - 2 17}{2 \cdot 1} : - 2 - 17

The solutions to the quadratic equation are:

c = - 2 + 17, c = - 2 - 17

4 x^{2} + 12 x + 9 = 0

5 x - 2 x^{2} = 2

s^{2} + 16 = 0

x^{2} - 5 x = - 20

2 p^{2} = 3 p

What are the solutions to the equation 6c^2+4c-13=5c^2 ?
The solutions to the equation 6c^2+4c-13=5c^2 are c=-2+sqrt(17),c=-2-sqrt(17)
Find the zeros of 6c^2+4c-13=5c^2
The zeros of 6c^2+4c-13=5c^2 are c=-2+sqrt(17),c=-2-sqrt(17)