2(x^2-5)=13x+35

Solution for 2(x^2-5)=13x+35 equation:

2(x^2-5)=13x+35

We move all terms to the left:

2(x^2-5)-(13x+35)=0

We multiply parentheses

2x^2-(13x+35)-10=0

We get rid of parentheses

2x^2-13x-35-10=0

We add all the numbers together, and all the variables

2x^2-13x-45=0

a = 2; b = -13; c = -45;
Δ = b2-4ac
Δ = -132-4·2·(-45)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-23}{2*2}=\frac{-10}{4}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+23}{2*2}=\frac{36}{4}
=9
$