4(3x2)=12x+8

Solution for 4(3x2)=12x+8 equation:

4(3x^2)=12x+8

We move all terms to the left:

4(3x^2)-(12x+8)=0

We get rid of parentheses

43x^2-12x-8=0

a = 43; b = -12; c = -8;
Δ = b2-4ac
Δ = -122-4·43·(-8)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{95}}{2*43}=\frac{12-4\sqrt{95}}{86}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{95}}{2*43}=\frac{12+4\sqrt{95}}{86}
$