What are the solutions to the equation 16x^2+kx-h=0 ?

The solutions to the equation 16x^2+kx-h=0 are x=(-k+sqrt(k^2+64h))/(32),x=(-k-sqrt(k^2+64h))/(32)

Find the zeros of 16x^2+kx-h=0

The zeros of 16x^2+kx-h=0 are x=(-k+sqrt(k^2+64h))/(32),x=(-k-sqrt(k^2+64h))/(32)

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

x = \frac{- k + k ^{2} + 64 h}{32}, x = \frac{- k - k ^{2} + 64 h}{32}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- k \pm k ^{2} - 4 \cdot 16 ( - h )}{2 \cdot 16}

Simplify

k^{2} - 4 \cdot 16 (- h) : k^{2} + 64 h

x_{1, 2} = \frac{- k \pm k ^{2} + 64 h}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- k + k ^{2} + 64 h}{2 \cdot 16}, x_{2} = \frac{- k - k ^{2} + 64 h}{2 \cdot 16}

x = \frac{- k + k ^{2} + 64 h}{2 \cdot 16} : \frac{- k + k ^{2} + 64 h}{32}

x = \frac{- k - k ^{2} + 64 h}{2 \cdot 16} : \frac{- k - k ^{2} + 64 h}{32}

The solutions to the quadratic equation are:

x = \frac{- k + k ^{2} + 64 h}{32}, x = \frac{- k - k ^{2} + 64 h}{32}

1 = 2 + x - x^{2}

r^{2} + 3 r - 54 = 0

x, f (a + b) = x^{2} - 5 x + 6

x^{2} - 5 x = 24

8 x^{2} = 5 x^{2} + 12

What are the solutions to the equation 16x^2+kx-h=0 ?
The solutions to the equation 16x^2+kx-h=0 are x=(-k+sqrt(k^2+64h))/(32),x=(-k-sqrt(k^2+64h))/(32)
Find the zeros of 16x^2+kx-h=0
The zeros of 16x^2+kx-h=0 are x=(-k+sqrt(k^2+64h))/(32),x=(-k-sqrt(k^2+64h))/(32)