4x^2+20x+25=48

Solution for 4x^2+20x+25=48 equation:

4x^2+20x+25=48

We move all terms to the left:

4x^2+20x+25-(48)=0

We add all the numbers together, and all the variables

4x^2+20x-23=0

a = 4; b = 20; c = -23;
Δ = b2-4ac
Δ = 202-4·4·(-23)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{3}}{2*4}=\frac{-20-16\sqrt{3}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{3}}{2*4}=\frac{-20+16\sqrt{3}}{8}
$