What are the solutions to the equation 2x^2-5x+(3-k)=0 ?

The solutions to the equation 2x^2-5x+(3-k)=0 are x=(5+sqrt(8k+1))/4 ,x=(5-sqrt(8k+1))/4

Find the zeros of 2x^2-5x+(3-k)=0

The zeros of 2x^2-5x+(3-k)=0 are x=(5+sqrt(8k+1))/4 ,x=(5-sqrt(8k+1))/4

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2 x^{2} - 5 x + (3 - k) = 0

x = \frac{5 + 8 k + 1}{4}, x = \frac{5 - 8 k + 1}{4}

Solution steps

2 x^{2} - 5 x + (3 - k) = 0

Refine

2 x^{2} - 5 x + 3 - k = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( 3 - k )}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 (3 - k) : 8 k + 1

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

Separate the solutions

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

x = \frac{- ( - 5 ) + 8 k + 1}{2 \cdot 2} : \frac{5 + 8 k + 1}{4}

x = \frac{- ( - 5 ) - 8 k + 1}{2 \cdot 2} : \frac{5 - 8 k + 1}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 8 k + 1}{4}, x = \frac{5 - 8 k + 1}{4}

x, 4 x^{2} + 6 x y + 4 y^{2} = 1

x^{2} - x (6 - 1) - 6 = 0

x, 2 x^{2} + p^{2} = (q - x)^{2}

a \cdot x^{2} - 8 x + 4 = 0

x, x^{2} y^{2} + x^{2} + y^{2} = 3736

What are the solutions to the equation 2x^2-5x+(3-k)=0 ?
The solutions to the equation 2x^2-5x+(3-k)=0 are x=(5+sqrt(8k+1))/4 ,x=(5-sqrt(8k+1))/4
Find the zeros of 2x^2-5x+(3-k)=0
The zeros of 2x^2-5x+(3-k)=0 are x=(5+sqrt(8k+1))/4 ,x=(5-sqrt(8k+1))/4