2x^2+14=68

Solution for 2x^2+14=68 equation:

2x^2+14=68

We move all terms to the left:

2x^2+14-(68)=0

We add all the numbers together, and all the variables

2x^2-54=0

a = 2; b = 0; c = -54;
Δ = b2-4ac
Δ = 02-4·2·(-54)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*2}=\frac{0-12\sqrt{3}}{4}
=-\frac{12\sqrt{3}}{4}
=-3\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*2}=\frac{0+12\sqrt{3}}{4}
=\frac{12\sqrt{3}}{4}
=3\sqrt{3}
$