4x-13+20x-10=x^2=1

Solution for 4x-13+20x-10=x^2=1 equation:

4x-13+20x-10=x^2=1

We move all terms to the left:

4x-13+20x-10-(x^2)=0

determiningTheFunctionDomain
-x^2+4x+20x-13-10=0

We add all the numbers together, and all the variables

-1x^2+24x-23=0

a = -1; b = 24; c = -23;
Δ = b2-4ac
Δ = 242-4·(-1)·(-23)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-22}{2*-1}=\frac{-46}{-2}
=+23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+22}{2*-1}=\frac{-2}{-2}
=1
$