2x^2-5=35

Solution for 2x^2-5=35 equation:

2x^2-5=35

We move all terms to the left:

2x^2-5-(35)=0

We add all the numbers together, and all the variables

2x^2-40=0

a = 2; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·2·(-40)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*2}=\frac{0-8\sqrt{5}}{4}
=-\frac{8\sqrt{5}}{4}
=-2\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*2}=\frac{0+8\sqrt{5}}{4}
=\frac{8\sqrt{5}}{4}
=2\sqrt{5}
$