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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16x^2+24x-187=0

a = 16; b = 24; c = -187;
Δ = b2-4ac
Δ = 242-4·16·(-187)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-112}{2*16}=\frac{-136}{32}
=-4+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+112}{2*16}=\frac{88}{32}
=2+3/4
$