import numpy as np

$ax^2 + bx + c = 0$

$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

a = np.float32(1.)
b = np.float32(-1000.001)
c = np.float32(1.)
x1 = (-b + np.sqrt(b * b - 4 * a * c, dtype=np.float32)) / (2 * a)
x2 = (-b - np.sqrt(b * b - 4 * a * c, dtype=np.float32)) / (2 * a)
print(x1, x2)

1000.0 0.001007080078125

print(a * x1 * x1 + b * x1 + c)
print(a * x2 * x2 + b * x2 + c)

0.0234375
-0.007080047391355038

$x_1 + x_2 = - \frac{b}{a}$

$x_1x_2 = \frac{c}{a}$

x1 = (-b + np.sqrt(b * b - 4 * a * c, dtype=np.float32)) / (2 * a)
print(x1)
x2 = c / x1 / a
x1 = -b/a - x2
print(x1, x2)
print(a * x1 * x1 + b * x1 + c)
print(a * x2 * x2 + b * x2 + c)

1000.0
999.9999765625 0.001
2.3399479687213898e-08
2.3437500051848303e-08

Let us write a little expressions for our roots

$$ \frac{1000.001 \pm \sqrt{1000.001^2 - 4}}{2} = \frac{1000.001 \pm 999.999}{2} = 1000 \text{ and } 0.00099999999$$

But if we have only 5 significant digits we have to truncate out exact answer to the form

$$ \frac{1000.001 \pm 999.999}{2} \approx \frac{1000.00 \pm 999.99}{2} = 999.95 \text{ and } 0.005$$

BUT!

$$ \frac{1}{999.95} = 0.00100005 $$

is much more accurate root!

$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

N = 1000000
items = np.array([1. / n**2 for n in range(1, N+1)], dtype=np.float32)
print(items.sum(), np.pi**2 / 6.)
print(np.abs(items.sum() - np.pi**2 / 6.))

1.6449317 1.6449340668482264
2.392844625331847e-06

val1 = np.float32(0.0)
for i in range(items.shape[0]):
    val1 += items[i]

print(val1, np.abs(val1 - np.pi**2 / 6.))

val2 = np.float32(0.0)
for i in range(items.shape[0]-1, -1, -1):
    val2 += items[i]

print(val2, np.abs(val2 - np.pi**2 / 6.))

1.6447253 0.0002087441248377342
1.644933 1.0815424402732532e-06