180-(3x^2+7x+1)=0

Solution for 180-(3x^2+7x+1)=0 equation:

180-(3x^2+7x+1)=0

We get rid of parentheses

-3x^2-7x-1+180=0

We add all the numbers together, and all the variables

-3x^2-7x+179=0

a = -3; b = -7; c = +179;
Δ = b2-4ac
Δ = -72-4·(-3)·179
Δ = 2197
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:

$\sqrt{\Delta}=\sqrt{2197}=\sqrt{169*13}=\sqrt{169}*\sqrt{13}=13\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13\sqrt{13}}{2*-3}=\frac{7-13\sqrt{13}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13\sqrt{13}}{2*-3}=\frac{7+13\sqrt{13}}{-6}
$