What are the solutions to the equation x^2-9x+c=0 ?

The solutions to the equation x^2-9x+c=0 are x=(9+sqrt(81-4c))/2 ,x=(9-sqrt(81-4c))/2

Find the zeros of x^2-9x+c=0

The zeros of x^2-9x+c=0 are x=(9+sqrt(81-4c))/2 ,x=(9-sqrt(81-4c))/2

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x^{2} - 9 x + c = 0

x = \frac{9 + 81 - 4 c}{2}, x = \frac{9 - 81 - 4 c}{2}

Solution steps

x^{2} - 9 x + c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot c}{2 \cdot 1}

Simplify

(- 9)^{2} - 4 \cdot 1 \cdot c : 81 - 4 c

x_{1, 2} = \frac{- ( - 9 ) \pm 81 - 4 c}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 81 - 4 c}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 81 - 4 c}{2 \cdot 1}

x = \frac{- ( - 9 ) + 81 - 4 c}{2 \cdot 1} : \frac{9 + 81 - 4 c}{2}

x = \frac{- ( - 9 ) - 81 - 4 c}{2 \cdot 1} : \frac{9 - 81 - 4 c}{2}

The solutions to the quadratic equation are:

x = \frac{9 + 81 - 4 c}{2}, x = \frac{9 - 81 - 4 c}{2}

x^{2} = 0.09

x^{2} - 4 x - 21 = 0

y, x^{2} + x y^{2} - y^{2} = 0

t, h = \frac{1}{2} g t^{2}

3 x^{2} + 39 = 30 x

What are the solutions to the equation x^2-9x+c=0 ?
The solutions to the equation x^2-9x+c=0 are x=(9+sqrt(81-4c))/2 ,x=(9-sqrt(81-4c))/2
Find the zeros of x^2-9x+c=0
The zeros of x^2-9x+c=0 are x=(9+sqrt(81-4c))/2 ,x=(9-sqrt(81-4c))/2