What are the solutions to the equation 7c^2-2c-15=-c^2 ?

The solutions to the equation 7c^2-2c-15=-c^2 are c= 3/2 ,c=-5/4

Find the zeros of 7c^2-2c-15=-c^2

The zeros of 7c^2-2c-15=-c^2 are c= 3/2 ,c=-5/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

7 c^{2} - 2 c - 15 = - c^{2}

c = \frac{3}{2}, c = - \frac{5}{4}

Decimal Notation

c = 1.5, c = - 1.25

Solution steps

7 c^{2} - 2 c - 15 = - c^{2}

Move

c^{2}

to the left side

8 c^{2} - 2 c - 15 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 15 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 15) = 22

c_{1, 2} = \frac{- ( - 2 ) \pm 22}{2 \cdot 8}

Separate the solutions

c_{1} = \frac{- ( - 2 ) + 22}{2 \cdot 8}, c_{2} = \frac{- ( - 2 ) - 22}{2 \cdot 8}

c = \frac{- ( - 2 ) + 22}{2 \cdot 8} : \frac{3}{2}

c = \frac{- ( - 2 ) - 22}{2 \cdot 8} : - \frac{5}{4}

The solutions to the quadratic equation are:

c = \frac{3}{2}, c = - \frac{5}{4}

3 a^{2} = 6 a - 3

3 x + 2 = x^{2} - 4 x - 28

28 - 3 = d - 7

∣ 2 x + 1 ∣ + 1 \leq 4

lo g_{9} (3) = w

What are the solutions to the equation 7c^2-2c-15=-c^2 ?
The solutions to the equation 7c^2-2c-15=-c^2 are c= 3/2 ,c=-5/4
Find the zeros of 7c^2-2c-15=-c^2
The zeros of 7c^2-2c-15=-c^2 are c= 3/2 ,c=-5/4