Question

$4y^2-80y+67=0$ solve it.

Answer 1

Given,

$4y^2-80y+67=0$

$a=4,b=-80,c=67$

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$y=\frac{-(-80)\pm \sqrt{{(-80)}^2-4\cdot 4\cdot 67}}{2\cdot 4}$

$=\frac{4(20\pm 3\sqrt{37})}{8}$

$\therefore y=\frac{20\pm 3\sqrt{37}}{2}$