3x^2-5+x=6x+1

Solution for 3x^2-5+x=6x+1 equation:

3x^2-5+x=6x+1

We move all terms to the left:

3x^2-5+x-(6x+1)=0

We get rid of parentheses

3x^2+x-6x-1-5=0

We add all the numbers together, and all the variables

3x^2-5x-6=0

a = 3; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·3·(-6)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{97}}{2*3}=\frac{5-\sqrt{97}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{97}}{2*3}=\frac{5+\sqrt{97}}{6}
$