What are the solutions to the equation c^2-15c+13=-6c ?

The solutions to the equation c^2-15c+13=-6c are c=(9+sqrt(29))/2 ,c=(9-sqrt(29))/2

Find the zeros of c^2-15c+13=-6c

The zeros of c^2-15c+13=-6c are c=(9+sqrt(29))/2 ,c=(9-sqrt(29))/2

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c^{2} - 15 c + 13 = - 6 c

c = \frac{9 + 29}{2}, c = \frac{9 - 29}{2}

Decimal Notation

c = 7.19258 \dots, c = 1.80741 \dots

Solution steps

c^{2} - 15 c + 13 = - 6 c

Move

6 c

to the left side

c^{2} - 9 c + 13 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 13 = 29

c_{1, 2} = \frac{- ( - 9 ) \pm 29}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- ( - 9 ) + 29}{2 \cdot 1}, c_{2} = \frac{- ( - 9 ) - 29}{2 \cdot 1}

c = \frac{- ( - 9 ) + 29}{2 \cdot 1} : \frac{9 + 29}{2}

c = \frac{- ( - 9 ) - 29}{2 \cdot 1} : \frac{9 - 29}{2}

The solutions to the quadratic equation are:

c = \frac{9 + 29}{2}, c = \frac{9 - 29}{2}

4 x + 20

- 3 \leq x \leq 6

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

∣ 6 ∣

- 30 = 5 (x + 1)

What are the solutions to the equation c^2-15c+13=-6c ?
The solutions to the equation c^2-15c+13=-6c are c=(9+sqrt(29))/2 ,c=(9-sqrt(29))/2
Find the zeros of c^2-15c+13=-6c
The zeros of c^2-15c+13=-6c are c=(9+sqrt(29))/2 ,c=(9-sqrt(29))/2