0=9x^2-62x+13

Solution for 0=9x^2-62x+13 equation:

0=9x^2-62x+13

We move all terms to the left:

0-(9x^2-62x+13)=0

We add all the numbers together, and all the variables

-(9x^2-62x+13)=0

We get rid of parentheses

-9x^2+62x-13=0

a = -9; b = 62; c = -13;
Δ = b2-4ac
Δ = 622-4·(-9)·(-13)
Δ = 3376
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:

$\sqrt{\Delta}=\sqrt{3376}=\sqrt{16*211}=\sqrt{16}*\sqrt{211}=4\sqrt{211}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-4\sqrt{211}}{2*-9}=\frac{-62-4\sqrt{211}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+4\sqrt{211}}{2*-9}=\frac{-62+4\sqrt{211}}{-18}
$