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

Solution for (4x^2-4x-3)=0 equation:

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

We get rid of parentheses

4x^2-4x-3=0

a = 4; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·4·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*4}=\frac{-4}{8}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*4}=\frac{12}{8}
=1+1/2
$