43=(3x)^2+7

Solution for 43=(3x)^2+7 equation:

43=(3x)^2+7

We move all terms to the left:

43-((3x)^2+7)=0

We get rid of parentheses

-3x^2-7+43=0

We add all the numbers together, and all the variables

-3x^2+36=0

a = -3; b = 0; c = +36;
Δ = b2-4ac
Δ = 02-4·(-3)·36
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*-3}=\frac{0-12\sqrt{3}}{-6}
=-\frac{12\sqrt{3}}{-6}
=-\frac{2\sqrt{3}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*-3}=\frac{0+12\sqrt{3}}{-6}
=\frac{12\sqrt{3}}{-6}
=\frac{2\sqrt{3}}{-1}
$