4(x^2-x+2)-5(x^2-2x+1)=0

Solution for 4(x^2-x+2)-5(x^2-2x+1)=0 equation:

4(x^2-x+2)-5(x^2-2x+1)=0

We multiply parentheses

4x^2-5x^2-4x+10x+8-5=0

We add all the numbers together, and all the variables

-1x^2+6x+3=0

a = -1; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·(-1)·3
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*-1}=\frac{-6-4\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*-1}=\frac{-6+4\sqrt{3}}{-2}
$