(4x+5)(4x+5)=125

Solution for (4x+5)(4x+5)=125 equation:

(4x+5)(4x+5)=125

We move all terms to the left:

(4x+5)(4x+5)-(125)=0

We multiply parentheses ..

(+16x^2+20x+20x+25)-125=0

We get rid of parentheses

16x^2+20x+20x+25-125=0

We add all the numbers together, and all the variables

16x^2+40x-100=0

a = 16; b = 40; c = -100;
Δ = b2-4ac
Δ = 402-4·16·(-100)
Δ = 8000
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:

$\sqrt{\Delta}=\sqrt{8000}=\sqrt{1600*5}=\sqrt{1600}*\sqrt{5}=40\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{5}}{2*16}=\frac{-40-40\sqrt{5}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{5}}{2*16}=\frac{-40+40\sqrt{5}}{32}
$