What are the solutions to the equation 2c^2+14c+10=-3 ?

The solutions to the equation 2c^2+14c+10=-3 are c=(-7+sqrt(23))/2 ,c=-(7+sqrt(23))/2

Find the zeros of 2c^2+14c+10=-3

The zeros of 2c^2+14c+10=-3 are c=(-7+sqrt(23))/2 ,c=-(7+sqrt(23))/2

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2 c^{2} + 14 c + 10 = - 3

c = \frac{- 7 + 23}{2}, c = - \frac{7 + 23}{2}

Decimal Notation

c = - 1.10208 \dots, c = - 5.89791 \dots

Solution steps

2 c^{2} + 14 c + 10 = - 3

Move

3

to the left side

2 c^{2} + 14 c + 13 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 \cdot 13}{2 \cdot 2}

1 4^{2} - 4 \cdot 2 \cdot 13 = 223

c_{1, 2} = \frac{- 14 \pm 2 23}{2 \cdot 2}

Separate the solutions

c_{1} = \frac{- 14 + 2 23}{2 \cdot 2}, c_{2} = \frac{- 14 - 2 23}{2 \cdot 2}

c = \frac{- 14 + 2 23}{2 \cdot 2} : \frac{- 7 + 23}{2}

c = \frac{- 14 - 2 23}{2 \cdot 2} : - \frac{7 + 23}{2}

The solutions to the quadratic equation are:

c = \frac{- 7 + 23}{2}, c = - \frac{7 + 23}{2}

r^{2} - 10 + 25 = 1

2 x^{2} - 6 = 11

2 x + 3 = x^{2} + x + 1

2 x^{2} + 2 x + 1

x^{2} + (x - 1)^{2} = 1

What are the solutions to the equation 2c^2+14c+10=-3 ?
The solutions to the equation 2c^2+14c+10=-3 are c=(-7+sqrt(23))/2 ,c=-(7+sqrt(23))/2
Find the zeros of 2c^2+14c+10=-3
The zeros of 2c^2+14c+10=-3 are c=(-7+sqrt(23))/2 ,c=-(7+sqrt(23))/2