Question

Solve $8x^2-10x-3=0$.

Answer 1

Given quadratic equation is $8x^2+10x-3=0$

We know that, for a quadratic equation of the form $a{x}^2+bx+c=0$,

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

Here, $a=8,b=-10andc=-3$

$\therefore x=\frac{-(-10)\pm \sqrt{(10{)}^2+4\times 8\times 3}}{2\times 8}$

$=\frac{10\pm \sqrt{100+96}}{16}$

$\Rightarrow x=\frac{10\pm 14}{16}$

$\Rightarrow x=\frac{10+14}{16}=\frac{24}{16}$

$\therefore x=\frac{3}{2}$

and $x=\frac{10-14}{16}=\frac{-4}{16}$

$\therefore x=\frac{-1}{4}$

Hence, the solution of the given quadratic equation is $x=\frac{3}{2}$ or $x=\frac{-1}{4}$.