X^2+3x+10+x+40=180

Solution for X^2+3x+10+x+40=180 equation:

X^2+3X+10+X+40=180

We move all terms to the left:

X^2+3X+10+X+40-(180)=0

We add all the numbers together, and all the variables

X^2+4X-130=0

a = 1; b = 4; c = -130;
Δ = b2-4ac
Δ = 42-4·1·(-130)
Δ = 536
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:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{134}}{2*1}=\frac{-4-2\sqrt{134}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{134}}{2*1}=\frac{-4+2\sqrt{134}}{2}
$