What are the solutions to the equation 16c^2-28c+9=-4c ?

The solutions to the equation 16c^2-28c+9=-4c are c= 3/4

Find the zeros of 16c^2-28c+9=-4c

The zeros of 16c^2-28c+9=-4c are c= 3/4

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16 c^{2} - 28 c + 9 = - 4 c

c = \frac{3}{4}

Solution steps

16 c^{2} - 28 c + 9 = - 4 c

Move

4 c

to the left side

16 c^{2} - 24 c + 9 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 16 \cdot 9}{2 \cdot 16}

(- 24)^{2} - 4 \cdot 16 \cdot 9 = 0

c_{1, 2} = \frac{- ( - 24 ) \pm 0}{2 \cdot 16}

c = \frac{- ( - 24 )}{2 \cdot 16}

\frac{- ( - 24 )}{2 \cdot 16} = \frac{3}{4}

c = \frac{3}{4}

The solution to the quadratic equation is:

c = \frac{3}{4}

(\frac{1}{9})^{- \frac{3}{2}}

(x - 1) (x + 2) (x + 3) > (x + 2) (x - 3)

- 10 < 2 + 2 x < 10

x^{2} + 2 x - 3 = 5

3 x = 96

What are the solutions to the equation 16c^2-28c+9=-4c ?
The solutions to the equation 16c^2-28c+9=-4c are c= 3/4
Find the zeros of 16c^2-28c+9=-4c
The zeros of 16c^2-28c+9=-4c are c= 3/4