Question

solve the quadratic equation


75
${75x }^{2} - 766x + 80 = 0$
​

Answer 1

Step-by-step explanation:

$75 {x}^{2} - 766x + 80 = 0$

To find roots of a quadratic equation

$\frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \: and \: \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$\frac{766 + \sqrt{ {(766)}^{2} - 4(75)(80)} }{2(75)} \\ = \frac{ \sqrt{140689} + 383}{75}$

$\frac{766 - \sqrt{ {(766)}^{2} - 4(75)(80)} }{2(75)} \\ = \frac{383 - \sqrt{140689} }{75}$