What are the solutions to the equation x^2-5x+c=0 ?

The solutions to the equation x^2-5x+c=0 are x=(5+sqrt(25-4c))/2 ,x=(5-sqrt(25-4c))/2

Find the zeros of x^2-5x+c=0

The zeros of x^2-5x+c=0 are x=(5+sqrt(25-4c))/2 ,x=(5-sqrt(25-4c))/2

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

x = \frac{5 + 25 - 4 c}{2}, x = \frac{5 - 25 - 4 c}{2}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot c}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot c : 25 - 4 c

x_{1, 2} = \frac{- ( - 5 ) \pm 25 - 4 c}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 25 - 4 c}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 25 - 4 c}{2 \cdot 1}

x = \frac{- ( - 5 ) + 25 - 4 c}{2 \cdot 1} : \frac{5 + 25 - 4 c}{2}

x = \frac{- ( - 5 ) - 25 - 4 c}{2 \cdot 1} : \frac{5 - 25 - 4 c}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 25 - 4 c}{2}, x = \frac{5 - 25 - 4 c}{2}

(1 - x) (2 - x) = 0

4 y^{2} - 1.6 y + 0.1225 = 0

f, 2 f + x (f)^{2} = 10

9^{2} = (62)^{2} + r^{2}

a, \frac{a ^{2} - b ^{2}}{9} = a + b

What are the solutions to the equation x^2-5x+c=0 ?
The solutions to the equation x^2-5x+c=0 are x=(5+sqrt(25-4c))/2 ,x=(5-sqrt(25-4c))/2
Find the zeros of x^2-5x+c=0
The zeros of x^2-5x+c=0 are x=(5+sqrt(25-4c))/2 ,x=(5-sqrt(25-4c))/2