What are the solutions to the equation 5x^2-6x+c=0 ?

The solutions to the equation 5x^2-6x+c=0 are x=(3+sqrt(-5c+9))/5 ,x=(3-sqrt(-5c+9))/5

Find the zeros of 5x^2-6x+c=0

The zeros of 5x^2-6x+c=0 are x=(3+sqrt(-5c+9))/5 ,x=(3-sqrt(-5c+9))/5

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5 x^{2} - 6 x + c = 0

x = \frac{3 + - 5 c + 9}{5}, x = \frac{3 - - 5 c + 9}{5}

Solution steps

5 x^{2} - 6 x + c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 5 c}{2 \cdot 5}

Simplify

(- 6)^{2} - 4 \cdot 5 c : 2 9 - 5 c

x_{1, 2} = \frac{- ( - 6 ) \pm 2 9 - 5 c}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 9 - 5 c}{2 \cdot 5}, x_{2} = \frac{- ( - 6 ) - 2 9 - 5 c}{2 \cdot 5}

x = \frac{- ( - 6 ) + 2 9 - 5 c}{2 \cdot 5} : \frac{3 + - 5 c + 9}{5}

x = \frac{- ( - 6 ) - 2 9 - 5 c}{2 \cdot 5} : \frac{3 - - 5 c + 9}{5}

The solutions to the quadratic equation are:

x = \frac{3 + - 5 c + 9}{5}, x = \frac{3 - - 5 c + 9}{5}

0 = 3 x^{2} - 4 x

21 x^{2} - 23 x - 20 = 0

3 p^{2} + 14 p - 1 = 6 p

(x + 1)^{2} + 3 = (x + 2)^{2} - x^{2}

2 x^{2} - 1 = x^{2} + 24

What are the solutions to the equation 5x^2-6x+c=0 ?
The solutions to the equation 5x^2-6x+c=0 are x=(3+sqrt(-5c+9))/5 ,x=(3-sqrt(-5c+9))/5
Find the zeros of 5x^2-6x+c=0
The zeros of 5x^2-6x+c=0 are x=(3+sqrt(-5c+9))/5 ,x=(3-sqrt(-5c+9))/5