20x^2+10-35=0

Solution for 20x^2+10-35=0 equation:

20x^2+10-35=0

We add all the numbers together, and all the variables

20x^2-25=0

a = 20; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·20·(-25)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*20}=\frac{0-20\sqrt{5}}{40}
=-\frac{20\sqrt{5}}{40}
=-\frac{\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*20}=\frac{0+20\sqrt{5}}{40}
=\frac{20\sqrt{5}}{40}
=\frac{\sqrt{5}}{2}
$