4x^2-80x+400=x+4

Solution for 4x^2-80x+400=x+4 equation:

4x^2-80x+400=x+4

We move all terms to the left:

4x^2-80x+400-(x+4)=0

We get rid of parentheses

4x^2-80x-x-4+400=0

We add all the numbers together, and all the variables

4x^2-81x+396=0

a = 4; b = -81; c = +396;
Δ = b2-4ac
Δ = -812-4·4·396
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-81)-15}{2*4}=\frac{66}{8}
=8+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-81)+15}{2*4}=\frac{96}{8}
=12
$