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

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

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

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

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

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

Solution steps

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}

3 x^{2} + 5 = 5 x^{2} - 8 x

m^{2} + m = 0

x^{2} + 64 = - 20 x

x^{2} - 2.5 x + 1.5 = 0

x^{2} - 19 x + 88 = 0

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