3x^2+6x-1=x^2+18x-14

Solution for 3x^2+6x-1=x^2+18x-14 equation:

3x^2+6x-1=x^2+18x-14

We move all terms to the left:

3x^2+6x-1-(x^2+18x-14)=0

We get rid of parentheses

3x^2-x^2+6x-18x+14-1=0

We add all the numbers together, and all the variables

2x^2-12x+13=0

a = 2; b = -12; c = +13;
Δ = b2-4ac
Δ = -122-4·2·13
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{10}}{2*2}=\frac{12-2\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{10}}{2*2}=\frac{12+2\sqrt{10}}{4}
$