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+1)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 24; c = -24;
Δ = b2-4ac
Δ = 242-4·(-1)·(-24)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{30}}{2*-1}=\frac{-24-4\sqrt{30}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{30}}{2*-1}=\frac{-24+4\sqrt{30}}{-2}
$