3x^2+6x-10=180

Solution for 3x^2+6x-10=180 equation:

3x^2+6x-10=180

We move all terms to the left:

3x^2+6x-10-(180)=0

We add all the numbers together, and all the variables

3x^2+6x-190=0

a = 3; b = 6; c = -190;
Δ = b2-4ac
Δ = 62-4·3·(-190)
Δ = 2316
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:

$\sqrt{\Delta}=\sqrt{2316}=\sqrt{4*579}=\sqrt{4}*\sqrt{579}=2\sqrt{579}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{579}}{2*3}=\frac{-6-2\sqrt{579}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{579}}{2*3}=\frac{-6+2\sqrt{579}}{6}
$