What are the solutions to the equation roots x^2+bx+c ?

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

Find the zeros of roots x^2+bx+c

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

roots

x^{2} + b x + c

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

Solution steps

x^{2} + b x + c

The roots are the intercepts with the x-axis

(y = 0)

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} + 2 x + 4

x, x^{2} + y^{2} + z^{2} - 4 x + 12 y - 14 z + 40 = 0

y, x^{2} + x y + y^{2} = 1

x^{2} - 5 x + 4 = 0

64 x^{2} - (16 a + 4 b) x + ab = 0

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