2x^2-13=20=(2x-5)(x-4)

Solution for 2x^2-13=20=(2x-5)(x-4) equation:

2x^2-13=20=(2x-5)(x-4)

We move all terms to the left:

2x^2-13-(20)=0

We add all the numbers together, and all the variables

2x^2-33=0

a = 2; b = 0; c = -33;
Δ = b2-4ac
Δ = 02-4·2·(-33)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{66}}{2*2}=\frac{0-2\sqrt{66}}{4}
=-\frac{2\sqrt{66}}{4}
=-\frac{\sqrt{66}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{66}}{2*2}=\frac{0+2\sqrt{66}}{4}
=\frac{2\sqrt{66}}{4}
=\frac{\sqrt{66}}{2}
$