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

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

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

We multiply parentheses

2x^2-10x+8=0

a = 2; b = -10; c = +8;
Δ = b2-4ac
Δ = -102-4·2·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*2}=\frac{16}{4}
=4
$