68x^2+1=16

Solution for 68x^2+1=16 equation:

68x^2+1=16

We move all terms to the left:

68x^2+1-(16)=0

We add all the numbers together, and all the variables

68x^2-15=0

a = 68; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·68·(-15)
Δ = 4080
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:

$\sqrt{\Delta}=\sqrt{4080}=\sqrt{16*255}=\sqrt{16}*\sqrt{255}=4\sqrt{255}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{255}}{2*68}=\frac{0-4\sqrt{255}}{136}
=-\frac{4\sqrt{255}}{136}
=-\frac{\sqrt{255}}{34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{255}}{2*68}=\frac{0+4\sqrt{255}}{136}
=\frac{4\sqrt{255}}{136}
=\frac{\sqrt{255}}{34}
$