What are the solutions to the equation 2x^2-hx+2k=0 ?

The solutions to the equation 2x^2-hx+2k=0 are x=(h+sqrt(h^2-16k))/4 ,x=(h-sqrt(h^2-16k))/4

Find the zeros of 2x^2-hx+2k=0

The zeros of 2x^2-hx+2k=0 are x=(h+sqrt(h^2-16k))/4 ,x=(h-sqrt(h^2-16k))/4

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2 x^{2} - h x + 2 k = 0

x = \frac{h + h ^{2} - 16 k}{4}, x = \frac{h - h ^{2} - 16 k}{4}

Solution steps

2 x^{2} - h x + 2 k = 0

Solve with the quadratic formula

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

Simplify

(- h)^{2} - 4 \cdot 2 \cdot 2 k : h^{2} - 16 k

x_{1, 2} = \frac{- ( - h ) \pm h ^{2} - 16 k}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - h ) + h ^{2} - 16 k}{2 \cdot 2}, x_{2} = \frac{- ( - h ) - h ^{2} - 16 k}{2 \cdot 2}

x = \frac{- ( - h ) + h ^{2} - 16 k}{2 \cdot 2} : \frac{h + h ^{2} - 16 k}{4}

x = \frac{- ( - h ) - h ^{2} - 16 k}{2 \cdot 2} : \frac{h - h ^{2} - 16 k}{4}

The solutions to the quadratic equation are:

x = \frac{h + h ^{2} - 16 k}{4}, x = \frac{h - h ^{2} - 16 k}{4}

3 v^{2} + 6 v + 3 = 0

r^{2} + 10 r + 50 = 0

49 + 16 m^{2} - 256 m = 0

- 4 + z^{2} = 16

2 x^{2} - 9 x = - 12

What are the solutions to the equation 2x^2-hx+2k=0 ?
The solutions to the equation 2x^2-hx+2k=0 are x=(h+sqrt(h^2-16k))/4 ,x=(h-sqrt(h^2-16k))/4
Find the zeros of 2x^2-hx+2k=0
The zeros of 2x^2-hx+2k=0 are x=(h+sqrt(h^2-16k))/4 ,x=(h-sqrt(h^2-16k))/4