What are the solutions to the equation 22c^2-35c+12=-3c^2 ?

The solutions to the equation 22c^2-35c+12=-3c^2 are c= 4/5 ,c= 3/5

Find the zeros of 22c^2-35c+12=-3c^2

The zeros of 22c^2-35c+12=-3c^2 are c= 4/5 ,c= 3/5

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22 c^{2} - 35 c + 12 = - 3 c^{2}

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

Decimal Notation

c = 0.8, c = 0.6

Solution steps

22 c^{2} - 35 c + 12 = - 3 c^{2}

Move

3 c^{2}

to the left side

25 c^{2} - 35 c + 12 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 35 ) \pm ( - 35 ) ^{2} - 4 \cdot 25 \cdot 12}{2 \cdot 25}

(- 35)^{2} - 4 \cdot 25 \cdot 12 = 5

c_{1, 2} = \frac{- ( - 35 ) \pm 5}{2 \cdot 25}

Separate the solutions

c_{1} = \frac{- ( - 35 ) + 5}{2 \cdot 25}, c_{2} = \frac{- ( - 35 ) - 5}{2 \cdot 25}

c = \frac{- ( - 35 ) + 5}{2 \cdot 25} : \frac{4}{5}

c = \frac{- ( - 35 ) - 5}{2 \cdot 25} : \frac{3}{5}

The solutions to the quadratic equation are:

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

z^{2} + 12 z = - 11

r^{2} + 22 r + 2 = 0

10 x^{2} - 148 x + 490 = 0

3 x^{2} - 240 = 0

4 x^{2} + 21 = 4 x

What are the solutions to the equation 22c^2-35c+12=-3c^2 ?
The solutions to the equation 22c^2-35c+12=-3c^2 are c= 4/5 ,c= 3/5
Find the zeros of 22c^2-35c+12=-3c^2
The zeros of 22c^2-35c+12=-3c^2 are c= 4/5 ,c= 3/5