15=2x^2-13x+36

Solution for 15=2x^2-13x+36 equation:

15=2x^2-13x+36

We move all terms to the left:

15-(2x^2-13x+36)=0

We get rid of parentheses

-2x^2+13x-36+15=0

We add all the numbers together, and all the variables

-2x^2+13x-21=0

a = -2; b = 13; c = -21;
Δ = b2-4ac
Δ = 132-4·(-2)·(-21)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*-2}=\frac{-14}{-4}
=3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*-2}=\frac{-12}{-4}
=+3
$