What are the solutions to the equation x^2+mx+n=0 ?

The solutions to the equation x^2+mx+n=0 are x=(-m+sqrt(m^2-4n))/2 ,x=(-m-sqrt(m^2-4n))/2

Find the zeros of x^2+mx+n=0

The zeros of x^2+mx+n=0 are x=(-m+sqrt(m^2-4n))/2 ,x=(-m-sqrt(m^2-4n))/2

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x^{2} + m x + n = 0

x = \frac{- m + m ^{2} - 4 n}{2}, x = \frac{- m - m ^{2} - 4 n}{2}

Solution steps

x^{2} + m x + n = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- m \pm m ^{2} - 4 \cdot 1 \cdot n}{2 \cdot 1}

Simplify

m^{2} - 4 \cdot 1 \cdot n : m^{2} - 4 n

x_{1, 2} = \frac{- m \pm m ^{2} - 4 n}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- m + m ^{2} - 4 n}{2 \cdot 1}, x_{2} = \frac{- m - m ^{2} - 4 n}{2 \cdot 1}

x = \frac{- m + m ^{2} - 4 n}{2 \cdot 1} : \frac{- m + m ^{2} - 4 n}{2}

x = \frac{- m - m ^{2} - 4 n}{2 \cdot 1} : \frac{- m - m ^{2} - 4 n}{2}

The solutions to the quadratic equation are:

x = \frac{- m + m ^{2} - 4 n}{2}, x = \frac{- m - m ^{2} - 4 n}{2}

10 - 7 s^{2} = 1

28 = - t^{2} + 3 t + 27

x^{2} - 8 x + 16 = 36

4^{2} + b^{2} = 7^{2}

1 0^{2} + y^{2} = 1 2^{2}

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