Question

2X^2-5x+2=0 Quadratic formula

Answer 1

Answer:

The solutions are $x_{1}=2\ and\ x_{2}=0.5$.

Step-by-step explanation:

Given:

$2x^{2} -5x+2=0$

Now, to solve by using quadratic formula:

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

Here $a=2,b=-5\ and\ c=2$

$x=\frac{-(-5)\pm \sqrt{(-5)^{2}-4\times 2\times 2}}{2\times 2}$

Now, here we will take either $x=\frac{-(-5)+\sqrt{25-16}}{4}$ and then $x=\frac{-(-5)-\sqrt{25-16}}{4}$.

$x=\frac{-(-5)+\sqrt{25-16}}{4}$        $x=\frac{-(-5)-\sqrt{25-16}}{4}$

$x=\frac{5+\sqrt{9}}{4}$               $x=\frac{5-\sqrt{9}}{4}$

$x=\frac{5+3}{4}$                         $x=\frac{5-3}{4}$

$x=\frac{8}{4}$                               $x=\frac{2}{4}$

$x_{1}=2$                                  $x_{2}=\frac{1}{2}=0.5$

Therefore, the solutions are $x_{1}=2\ and\ x_{2}=0.5$.