What are the solutions to the equation x^2+cx+b=0 ?

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

Find the zeros of x^2+cx+b=0

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

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x, lo g_{10} (5) x^{2} - lo g_{10} (5 y) = 1

32 y^{2} + 32 y + 8 = 0

x^{2} = \frac{lo g _{10} ( 25 )}{lo g _{10} ( 2 )}

x, x^{2} - 2 x y - y^{2} + x - 12 y = 0

x, 16 x^{2} - y^{2} + 16 z^{2} = 16

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