4(x^2-3x-1)=0

Solution for 4(x^2-3x-1)=0 equation:

4(x^2-3x-1)=0

We multiply parentheses

4x^2-12x-4=0

a = 4; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·4·(-4)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{13}}{2*4}=\frac{12-4\sqrt{13}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{13}}{2*4}=\frac{12+4\sqrt{13}}{8}
$