(2x)^2+15=17^2

Solution for (2x)^2+15=17^2 equation:

(2x)^2+15=17^2

We move all terms to the left:

(2x)^2+15-(17^2)=0

We add all the numbers together, and all the variables

2x^2-274=0

a = 2; b = 0; c = -274;
Δ = b2-4ac
Δ = 02-4·2·(-274)
Δ = 2192
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:

$\sqrt{\Delta}=\sqrt{2192}=\sqrt{16*137}=\sqrt{16}*\sqrt{137}=4\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{137}}{2*2}=\frac{0-4\sqrt{137}}{4}
=-\frac{4\sqrt{137}}{4}
=-\sqrt{137}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{137}}{2*2}=\frac{0+4\sqrt{137}}{4}
=\frac{4\sqrt{137}}{4}
=\sqrt{137}
$