(6x^2+17)-(3x^2+20)=0

Solution for (6x^2+17)-(3x^2+20)=0 equation:

(6x^2+17)-(3x^2+20)=0

We get rid of parentheses

6x^2-3x^2+17-20=0

We add all the numbers together, and all the variables

3x^2-3=0

a = 3; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·3·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*3}=\frac{-6}{6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*3}=\frac{6}{6}
=1
$