(2x+1)(2x+3)=195

Solution for (2x+1)(2x+3)=195 equation:

(2x+1)(2x+3)=195

We move all terms to the left:

(2x+1)(2x+3)-(195)=0

We multiply parentheses ..

(+4x^2+6x+2x+3)-195=0

We get rid of parentheses

4x^2+6x+2x+3-195=0

We add all the numbers together, and all the variables

4x^2+8x-192=0

a = 4; b = 8; c = -192;
Δ = b2-4ac
Δ = 82-4·4·(-192)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-56}{2*4}=\frac{-64}{8}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+56}{2*4}=\frac{48}{8}
=6
$