4x^2+20x+25=16

Solution for 4x^2+20x+25=16 equation:

4x^2+20x+25=16

We move all terms to the left:

4x^2+20x+25-(16)=0

We add all the numbers together, and all the variables

4x^2+20x+9=0

a = 4; b = 20; c = +9;
Δ = b2-4ac
Δ = 202-4·4·9
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{-(20)-16}{2*4}=\frac{-36}{8}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*4}=\frac{-4}{8}
=-1/2
$