10-10x+2.5x^2=5

Solution for 10-10x+2.5x^2=5 equation:

10-10x+2.5x^2=5

We move all terms to the left:

10-10x+2.5x^2-(5)=0

We add all the numbers together, and all the variables

2.5x^2-10x+5=0

a = 2.5; b = -10; c = +5;
Δ = b2-4ac
Δ = -102-4·2.5·5
Δ = 50
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:

$\sqrt{\Delta}=\sqrt{50}=\sqrt{25*2}=\sqrt{25}*\sqrt{2}=5\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-5\sqrt{2}}{2*2.5}=\frac{10-5\sqrt{2}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+5\sqrt{2}}{2*2.5}=\frac{10+5\sqrt{2}}{5}
$