(2x^2-4x+8)-(x^2+3x-1)=1

Solution for (2x^2-4x+8)-(x^2+3x-1)=1 equation:

(2x^2-4x+8)-(x^2+3x-1)=1

We move all terms to the left:

(2x^2-4x+8)-(x^2+3x-1)-(1)=0

We get rid of parentheses

2x^2-x^2-4x-3x+8+1-1=0

We add all the numbers together, and all the variables

x^2-7x+8=0

a = 1; b = -7; c = +8;
Δ = b2-4ac
Δ = -72-4·1·8
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{17}}{2*1}=\frac{7-\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{17}}{2*1}=\frac{7+\sqrt{17}}{2}
$