We are given this equation:
![2x^{2} -12x+20=0](https://tex.z-dn.net/?f=%202x%5E%7B2%7D%20-12x%2B20%3D0%20)
Let us use quadratic formula to find its solution.
The quadratic formula for general quadratic equation of the form ax²+bx+c=0 is given by:
![x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%7D%7B2a%7D)
Now on comparing our equation with this general form our a, b and c are:
a=2, b=-12 and c=20
Plugging these values in the formula:
![x=\frac{-(-12))\pm \sqrt{(-12)^{2}-4(2)(20)}}{2(2))}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-12%29%29%5Cpm%20%5Csqrt%7B%28-12%29%5E%7B2%7D-4%282%29%2820%29%7D%7D%7B2%282%29%29%7D)
On solving we will get,
![x=\frac{(12))\pm \sqrt{(144)-(160)}}{(4)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2812%29%29%5Cpm%20%5Csqrt%7B%28144%29-%28160%29%7D%7D%7B%284%29%7D)
Simplifying we will get,
![x=\frac{(12))\pm \sqrt{-16}}{(4)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2812%29%29%5Cpm%20%5Csqrt%7B-16%7D%7D%7B%284%29%7D)
square root of -1 is "i".
![x=\frac{(12))\pm \-4i}{(4)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2812%29%29%5Cpm%20%5C-4i%7D%7B%284%29%7D)
taking 4 common out as gcf in numerator,
![x=3\pm i](https://tex.z-dn.net/?f=%20x%3D3%5Cpm%20i%20)
Answer : x=3+i and x=3-i