(x^2-5)+(4x-2)=90

Solution for (x^2-5)+(4x-2)=90 equation:

(x^2-5)+(4x-2)=90

We move all terms to the left:

(x^2-5)+(4x-2)-(90)=0

We get rid of parentheses

x^2+4x-5-2-90=0

We add all the numbers together, and all the variables

x^2+4x-97=0

a = 1; b = 4; c = -97;
Δ = b2-4ac
Δ = 42-4·1·(-97)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{101}}{2*1}=\frac{-4-2\sqrt{101}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{101}}{2*1}=\frac{-4+2\sqrt{101}}{2}
$