What are the solutions to the equation 1x^2+bx+c=0 ?

The solutions to the equation 1x^2+bx+c=0 are x=(-b+sqrt(b^2-4c))/2 ,x=(-b-sqrt(b^2-4c))/2

Find the zeros of 1x^2+bx+c=0

The zeros of 1x^2+bx+c=0 are x=(-b+sqrt(b^2-4c))/2 ,x=(-b-sqrt(b^2-4c))/2

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

x = \frac{- b + b ^{2} - 4 c}{2}, x = \frac{- b - b ^{2} - 4 c}{2}

Solution steps

1 \cdot x^{2} + b x + c = 0

Refine

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

Solve with the quadratic formula

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

Simplify

b^{2} - 4 \cdot 1 \cdot c : b^{2} - 4 c

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

Separate the solutions

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

x = \frac{- b + b ^{2} - 4 c}{2 \cdot 1} : \frac{- b + b ^{2} - 4 c}{2}

x = \frac{- b - b ^{2} - 4 c}{2 \cdot 1} : \frac{- b - b ^{2} - 4 c}{2}

The solutions to the quadratic equation are:

x = \frac{- b + b ^{2} - 4 c}{2}, x = \frac{- b - b ^{2} - 4 c}{2}

x^{2} + x + 13 = 4 + 7 x

12 y^{2} = - 5 y^{2} + 2

(x - 4) (x - 6) = - 1

3 x^{2} + 9 x

x, 10 x^{2} - 8 y - 8 y^{3} = 0

What are the solutions to the equation 1x^2+bx+c=0 ?
The solutions to the equation 1x^2+bx+c=0 are x=(-b+sqrt(b^2-4c))/2 ,x=(-b-sqrt(b^2-4c))/2
Find the zeros of 1x^2+bx+c=0
The zeros of 1x^2+bx+c=0 are x=(-b+sqrt(b^2-4c))/2 ,x=(-b-sqrt(b^2-4c))/2