What are the solutions to the equation c^2+c-19=4c ?

The solutions to the equation c^2+c-19=4c are c=(3+sqrt(85))/2 ,c=(3-sqrt(85))/2

Find the zeros of c^2+c-19=4c

The zeros of c^2+c-19=4c are c=(3+sqrt(85))/2 ,c=(3-sqrt(85))/2

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c^{2} + c - 19 = 4 c

c = \frac{3 + 85}{2}, c = \frac{3 - 85}{2}

Decimal Notation

c = 6.10977 \dots, c = - 3.10977 \dots

Solution steps

c^{2} + c - 19 = 4 c

Move

4 c

to the left side

c^{2} - 3 c - 19 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 19) = 85

c_{1, 2} = \frac{- ( - 3 ) \pm 85}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- ( - 3 ) + 85}{2 \cdot 1}, c_{2} = \frac{- ( - 3 ) - 85}{2 \cdot 1}

c = \frac{- ( - 3 ) + 85}{2 \cdot 1} : \frac{3 + 85}{2}

c = \frac{- ( - 3 ) - 85}{2 \cdot 1} : \frac{3 - 85}{2}

The solutions to the quadratic equation are:

c = \frac{3 + 85}{2}, c = \frac{3 - 85}{2}

- x^{2} + 6 x = 10

y, x^{2} - x (y - 2) - 3 y = 0

x^{2} - \frac{11}{10} x + \frac{1}{10} = 0

A, A = 2 π r^{2} + 2 π

4 u^{2} + 20 u = - 25

What are the solutions to the equation c^2+c-19=4c ?
The solutions to the equation c^2+c-19=4c are c=(3+sqrt(85))/2 ,c=(3-sqrt(85))/2
Find the zeros of c^2+c-19=4c
The zeros of c^2+c-19=4c are c=(3+sqrt(85))/2 ,c=(3-sqrt(85))/2