How to solve 5 x 2 - 3 x = 25 by completing the square?

So solution:
${\displaystyle 5x^2-3x=25\text{or}5(x^2-\frac{3}{5}x)=25}$
${\displaystyle 5(x^2-\frac{3}{5}x+\frac{9}{100})=25+\frac{9}{20}}$
${\displaystyle 5{(x-\frac{3}{10})}^2=\frac{509}{20}}$
${\displaystyle {(x-\frac{3}{10})}^2=\frac{509}{100}}$
${\displaystyle {(x-\frac{3}{10})}^2=\frac{509}{100}}$
${\displaystyle x=0.3\pm 2.256\therefore x=2.556,x=-1.956}$
Result: ${\displaystyle x\approx 2.556,x\approx -1.956}$

Here is an example of how to solve using the difference of squares identity, premultiplying by 20 to reduce the number of fractions, and completing the square:
${\displaystyle a^2-b^2=(a-b)(a+b)}$
with #a=(10x-3)# and #b=sqrt(509)#
We have:
${\displaystyle 5x^2-3x=25}$
25 from each side is subtracted to yield:
${\displaystyle 5x^2-3x-25=0}$
Multiply by ${\displaystyle 20=5\cdot 2^2}$ to make the leading term a perfect square and the middle term divisible by 2...
Therefore:
${\displaystyle 0=20(5x^2-3x-25)}$
${\displaystyle 0=100x^2-60x-500}$
${\displaystyle 0={\left(10x\right)}^2-2\left(10x\right)\left(3\right)+3^2-509}$
${\displaystyle 0={(10x-3)}^2-{\left(\sqrt{509}\right)}^2}$
${\displaystyle 0=((10x-3)-\sqrt{509})((10x-3)+\sqrt{509})}$
${\displaystyle 0=(10x-3-\sqrt{509})(10x-3+\sqrt{509})}$
Hence:
${\displaystyle 10x=3\pm \sqrt{509}}$
So:
${\displaystyle x=\frac{3}{10}\pm \frac{\sqrt{509}}{10}}$