How do you solve $10 x^{2} + 9 x = 499$ ?

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Solution

$x 1 = 7.53$ $x 2 = - 6.63$

Explanation:

$y = 10 x^{2} + 9 x = 499$

Use the improved quadratic formula (Socratic search):

$D = d^{2} = b^{2} - 4 a c = 81 + 19, 960 = 20, 041$

$d = \pm 141.57$

There are 2 real roots:

$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{9}{20} \pm \frac{141.57}{20}$

$x = \frac{9 \pm 141.57}{20}$

$x 1 = \frac{150.57}{20} = 7.53$

$x 2 = - \frac{132.57}{20} = - 6.63$

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