3x^2+1-7x=13x^

Solution for 3x^2+1-7x=13x^ equation:

3x^2+1-7x=13x^

We move all terms to the left:

3x^2+1-7x-(13x^)=0

We add all the numbers together, and all the variables

3x^2-20x+1=0

a = 3; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·3·1
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{97}}{2*3}=\frac{20-2\sqrt{97}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{97}}{2*3}=\frac{20+2\sqrt{97}}{6}
$