h(2)=16(2)^2+48(2)+4

Solution for h(2)=16(2)^2+48(2)+4 equation:

h(2)=16(2)^2+48(2)+4

We move all terms to the left:

h(2)-(16(2)^2+48(2)+4)=0

We add all the numbers together, and all the variables

h^2-26730=0

a = 1; b = 0; c = -26730;
Δ = b2-4ac
Δ = 02-4·1·(-26730)
Δ = 106920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{106920}=\sqrt{324*330}=\sqrt{324}*\sqrt{330}=18\sqrt{330}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{330}}{2*1}=\frac{0-18\sqrt{330}}{2}
=-\frac{18\sqrt{330}}{2}
=-9\sqrt{330}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{330}}{2*1}=\frac{0+18\sqrt{330}}{2}
=\frac{18\sqrt{330}}{2}
=9\sqrt{330}
$