Can someone help me please

(2x-10)(x+5)=x^2+10x-75

okay:  x^2-10x=-25

now: $$x^2-10x+25=0$$

use the abc - formula:

if $$ax^2+bx+c = 0$$  than $$x_{1,2} = \frac{1}{2a}*(b-\sqrt{b^2-4*a*c})$$

$$1x^2 -10x+25=0 \\ \underbrace{1}_{a=1}x^2 \underbrace{-10}_{ b=-10}x+\underbrace{25}_{c=25}=0 \\\\ x_{1,2} = \frac{1}{2*1}*(10-\sqrt{100-4*1*25})\\\\ x_{1,2} = \frac{1}{2}*(10-\sqrt{100-100})\\\\ x = \frac{1}{2}*(10-0)\\\\ x = \frac{1}{2}*10\\\\ x = \frac{10}{2}\\\\ x = 5$$

Great job Heureka!  You're smart!

and can i solve it this way?

x^2-10x+25

(x-5)^2=0

√(x-5)^2=√0

x-5=0

x=5