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

The solutions to the equation 2x^2-5x+c=0 are x=(5+sqrt(25-8c))/4 ,x=(5-sqrt(25-8c))/4

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

The zeros of 2x^2-5x+c=0 are x=(5+sqrt(25-8c))/4 ,x=(5-sqrt(25-8c))/4

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

x = \frac{5 + 25 - 8 c}{4}, x = \frac{5 - 25 - 8 c}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 2 c : 25 - 8 c

x_{1, 2} = \frac{- ( - 5 ) \pm 25 - 8 c}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 25 - 8 c}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 25 - 8 c}{2 \cdot 2}

x = \frac{- ( - 5 ) + 25 - 8 c}{2 \cdot 2} : \frac{5 + 25 - 8 c}{4}

x = \frac{- ( - 5 ) - 25 - 8 c}{2 \cdot 2} : \frac{5 - 25 - 8 c}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 25 - 8 c}{4}, x = \frac{5 - 25 - 8 c}{4}

3 X^{2} = 5 X

8 x^{2} + 22 x = - 15

x, 1 = x^{2} + y

8^{2} = a^{2} - 6^{2}

x, 0 = 0.81 x^{2} - 11.78 x + 13.52

What are the solutions to the equation 2x^2-5x+c=0 ?
The solutions to the equation 2x^2-5x+c=0 are x=(5+sqrt(25-8c))/4 ,x=(5-sqrt(25-8c))/4
Find the zeros of 2x^2-5x+c=0
The zeros of 2x^2-5x+c=0 are x=(5+sqrt(25-8c))/4 ,x=(5-sqrt(25-8c))/4