4(3x2)=68

Solution for 4(3x2)=68 equation:

4(3x^2)=68

We move all terms to the left:

4(3x^2)-(68)=0

a = 43; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·43·(-68)
Δ = 11696
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:

$\sqrt{\Delta}=\sqrt{11696}=\sqrt{16*731}=\sqrt{16}*\sqrt{731}=4\sqrt{731}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{731}}{2*43}=\frac{0-4\sqrt{731}}{86}
=-\frac{4\sqrt{731}}{86}
=-\frac{2\sqrt{731}}{43}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{731}}{2*43}=\frac{0+4\sqrt{731}}{86}
=\frac{4\sqrt{731}}{86}
=\frac{2\sqrt{731}}{43}
$