What are the solutions to the equation 7x^2+bx+c=0 ?

The solutions to the equation 7x^2+bx+c=0 are x=(-b+sqrt(b^2-28c))/(14),x=(-b-sqrt(b^2-28c))/(14)

Find the zeros of 7x^2+bx+c=0

The zeros of 7x^2+bx+c=0 are x=(-b+sqrt(b^2-28c))/(14),x=(-b-sqrt(b^2-28c))/(14)

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7 x^{2} + b x + c = 0

x = \frac{- b + b ^{2} - 28 c}{14}, x = \frac{- b - b ^{2} - 28 c}{14}

Solution steps

7 x^{2} + b x + c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- b \pm b ^{2} - 4 \cdot 7 c}{2 \cdot 7}

Simplify

b^{2} - 4 \cdot 7 c : b^{2} - 28 c

x_{1, 2} = \frac{- b \pm b ^{2} - 28 c}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- b + b ^{2} - 28 c}{2 \cdot 7}, x_{2} = \frac{- b - b ^{2} - 28 c}{2 \cdot 7}

x = \frac{- b + b ^{2} - 28 c}{2 \cdot 7} : \frac{- b + b ^{2} - 28 c}{14}

x = \frac{- b - b ^{2} - 28 c}{2 \cdot 7} : \frac{- b - b ^{2} - 28 c}{14}

The solutions to the quadratic equation are:

x = \frac{- b + b ^{2} - 28 c}{14}, x = \frac{- b - b ^{2} - 28 c}{14}

5 v^{2} + 30 v = 0

3 x^{2} + x - 7 = 0

m^{2} - 2 m - 5 = 0

3 x^{2} - 4 x = 5

x^{2} - 4 = 32

What are the solutions to the equation 7x^2+bx+c=0 ?
The solutions to the equation 7x^2+bx+c=0 are x=(-b+sqrt(b^2-28c))/(14),x=(-b-sqrt(b^2-28c))/(14)
Find the zeros of 7x^2+bx+c=0
The zeros of 7x^2+bx+c=0 are x=(-b+sqrt(b^2-28c))/(14),x=(-b-sqrt(b^2-28c))/(14)