What are the solutions to the equation x^2+mx+7=0,m>0 ?

The solutions to the equation x^2+mx+7=0,m>0 are x=(-m+sqrt(m^2-28))/2 ,x=(-m-sqrt(m^2-28))/2

Find the zeros of x^2+mx+7=0,m>0

The zeros of x^2+mx+7=0,m>0 are x=(-m+sqrt(m^2-28))/2 ,x=(-m-sqrt(m^2-28))/2

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x^{2} + m x + 7 = 0, m > 0

x = \frac{- m + m ^{2} - 28}{2}, x = \frac{- m - m ^{2} - 28}{2}

Solution steps

x^{2} + m x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- m \pm m ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

Simplify

m^{2} - 4 \cdot 1 \cdot 7 : m^{2} - 28

x_{1, 2} = \frac{- m \pm m ^{2} - 28}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- m + m ^{2} - 28}{2 \cdot 1}, x_{2} = \frac{- m - m ^{2} - 28}{2 \cdot 1}

x = \frac{- m + m ^{2} - 28}{2 \cdot 1} : \frac{- m + m ^{2} - 28}{2}

x = \frac{- m - m ^{2} - 28}{2 \cdot 1} : \frac{- m - m ^{2} - 28}{2}

The solutions to the quadratic equation are:

x = \frac{- m + m ^{2} - 28}{2}, x = \frac{- m - m ^{2} - 28}{2}

3 x^{2} - x = - 2

2 y^{2} = 6 y

- 7 x - 4 = - 3 x^{2}

s^{2} + s + 4 = 0

3 x^{2} + 10 x - 17 = - 9

What are the solutions to the equation x^2+mx+7=0,m>0 ?
The solutions to the equation x^2+mx+7=0,m>0 are x=(-m+sqrt(m^2-28))/2 ,x=(-m-sqrt(m^2-28))/2
Find the zeros of x^2+mx+7=0,m>0
The zeros of x^2+mx+7=0,m>0 are x=(-m+sqrt(m^2-28))/2 ,x=(-m-sqrt(m^2-28))/2