4(3x2)=112

Solution for 4(3x2)=112 equation:

4(3x^2)=112

We move all terms to the left:

4(3x^2)-(112)=0

a = 43; b = 0; c = -112;
Δ = b2-4ac
Δ = 02-4·43·(-112)
Δ = 19264
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:

$\sqrt{\Delta}=\sqrt{19264}=\sqrt{64*301}=\sqrt{64}*\sqrt{301}=8\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{301}}{2*43}=\frac{0-8\sqrt{301}}{86}
=-\frac{8\sqrt{301}}{86}
=-\frac{4\sqrt{301}}{43}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{301}}{2*43}=\frac{0+8\sqrt{301}}{86}
=\frac{8\sqrt{301}}{86}
=\frac{4\sqrt{301}}{43}
$