What are the solutions to the equation roots x^2+ax+b ?

The solutions to the equation roots x^2+ax+b are x=(-a+sqrt(a^2-4b))/2 ,x=(-a-sqrt(a^2-4b))/2

Find the zeros of roots x^2+ax+b

The zeros of roots x^2+ax+b are x=(-a+sqrt(a^2-4b))/2 ,x=(-a-sqrt(a^2-4b))/2

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roots

x^{2} + a x + b

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

Solution steps

x^{2} + a x + b

The roots are the intercepts with the x-axis

(y = 0)

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{d}{d x} (4 cos (π x))

\frac{693 x ^{\frac{5}{2}}}{8}

(x, y) \to (0, 0) lim (x y)

\frac{\partial}{\partial x} (e^{- (x - v t)^{2}})

\frac{\partial}{\partial x} (x cos (x) sin (x))

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