(2x+24)+(4x+43)+(x2+1)=180

Solution for (2x+24)+(4x+43)+(x2+1)=180 equation:

(2x+24)+(4x+43)+(x2+1)=180

We move all terms to the left:

(2x+24)+(4x+43)+(x2+1)-(180)=0

We add all the numbers together, and all the variables

(+x^2+1)+(2x+24)+(4x+43)-180=0

We get rid of parentheses

x^2+2x+4x+1+24+43-180=0

We add all the numbers together, and all the variables

x^2+6x-112=0

a = 1; b = 6; c = -112;
Δ = b2-4ac
Δ = 62-4·1·(-112)
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{-(6)-22}{2*1}=\frac{-28}{2}
=-14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+22}{2*1}=\frac{16}{2}
=8
$