What are the solutions to the equation 15c^2-16c+3=-c^2 ?

The solutions to the equation 15c^2-16c+3=-c^2 are c= 3/4 ,c= 1/4

Find the zeros of 15c^2-16c+3=-c^2

The zeros of 15c^2-16c+3=-c^2 are c= 3/4 ,c= 1/4

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15 c^{2} - 16 c + 3 = - c^{2}

c = \frac{3}{4}, c = \frac{1}{4}

Decimal Notation

c = 0.75, c = 0.25

Solution steps

15 c^{2} - 16 c + 3 = - c^{2}

Move

c^{2}

to the left side

16 c^{2} - 16 c + 3 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 16 \cdot 3}{2 \cdot 16}

(- 16)^{2} - 4 \cdot 16 \cdot 3 = 8

c_{1, 2} = \frac{- ( - 16 ) \pm 8}{2 \cdot 16}

Separate the solutions

c_{1} = \frac{- ( - 16 ) + 8}{2 \cdot 16}, c_{2} = \frac{- ( - 16 ) - 8}{2 \cdot 16}

c = \frac{- ( - 16 ) + 8}{2 \cdot 16} : \frac{3}{4}

c = \frac{- ( - 16 ) - 8}{2 \cdot 16} : \frac{1}{4}

The solutions to the quadratic equation are:

c = \frac{3}{4}, c = \frac{1}{4}

2 - 5

(8 a^{2} z) (\frac{1}{2} a^{3} z^{4})

7 - 5 + 3 x - 1 = 7 + 3

7 \cdot 10

\frac{5}{2} + 4

What are the solutions to the equation 15c^2-16c+3=-c^2 ?
The solutions to the equation 15c^2-16c+3=-c^2 are c= 3/4 ,c= 1/4
Find the zeros of 15c^2-16c+3=-c^2
The zeros of 15c^2-16c+3=-c^2 are c= 3/4 ,c= 1/4