2x(x-5)+10=0

Solution for 2x(x-5)+10=0 equation:

2x(x-5)+10=0

We multiply parentheses

2x^2-10x+10=0

a = 2; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·2·10
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{5}}{2*2}=\frac{10-2\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*2}=\frac{10+2\sqrt{5}}{4}
$