(4x+3)(4x+3)=196+4x^2

Solution for (4x+3)(4x+3)=196+4x^2 equation:

(4x+3)(4x+3)=196+4x^2

We move all terms to the left:

(4x+3)(4x+3)-(196+4x^2)=0

We get rid of parentheses

-4x^2+(4x+3)(4x+3)-196=0

We multiply parentheses ..

-4x^2+(+16x^2+12x+12x+9)-196=0

We get rid of parentheses

-4x^2+16x^2+12x+12x+9-196=0

We add all the numbers together, and all the variables

12x^2+24x-187=0

a = 12; b = 24; c = -187;
Δ = b2-4ac
Δ = 242-4·12·(-187)
Δ = 9552
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:

$\sqrt{\Delta}=\sqrt{9552}=\sqrt{16*597}=\sqrt{16}*\sqrt{597}=4\sqrt{597}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{597}}{2*12}=\frac{-24-4\sqrt{597}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{597}}{2*12}=\frac{-24+4\sqrt{597}}{24}
$