16x^2-160x+400=4

Solution for 16x^2-160x+400=4 equation:

16x^2-160x+400=4

We move all terms to the left:

16x^2-160x+400-(4)=0

We add all the numbers together, and all the variables

16x^2-160x+396=0

a = 16; b = -160; c = +396;
Δ = b2-4ac
Δ = -1602-4·16·396
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-16}{2*16}=\frac{144}{32}
=4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+16}{2*16}=\frac{176}{32}
=5+1/2
$