2w^2+4-w^2-20w+25=0

Solution for 2w^2+4-w^2-20w+25=0 equation:

2w^2+4-w^2-20w+25=0

We add all the numbers together, and all the variables

w^2-20w+29=0

a = 1; b = -20; c = +29;
Δ = b2-4ac
Δ = -202-4·1·29
Δ = 284
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{71}}{2*1}=\frac{20-2\sqrt{71}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{71}}{2*1}=\frac{20+2\sqrt{71}}{2}
$