7(6^2x+3)=1512

Solution for 7(6^2x+3)=1512 equation:

7(6^2x+3)=1512

We move all terms to the left:

7(6^2x+3)-(1512)=0

We multiply parentheses

42x^2+21-1512=0

We add all the numbers together, and all the variables

42x^2-1491=0

a = 42; b = 0; c = -1491;
Δ = b2-4ac
Δ = 02-4·42·(-1491)
Δ = 250488
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:

$\sqrt{\Delta}=\sqrt{250488}=\sqrt{1764*142}=\sqrt{1764}*\sqrt{142}=42\sqrt{142}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{142}}{2*42}=\frac{0-42\sqrt{142}}{84}
=-\frac{42\sqrt{142}}{84}
=-\frac{\sqrt{142}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{142}}{2*42}=\frac{0+42\sqrt{142}}{84}
=\frac{42\sqrt{142}}{84}
=\frac{\sqrt{142}}{2}
$