x^2+3x+50=180-4x-10

Solution for x^2+3x+50=180-4x-10 equation:

x^2+3x+50=180-4x-10

We move all terms to the left:

x^2+3x+50-(180-4x-10)=0

We add all the numbers together, and all the variables

x^2+3x-(-4x+170)+50=0

We get rid of parentheses

x^2+3x+4x-170+50=0

We add all the numbers together, and all the variables

x^2+7x-120=0

a = 1; b = 7; c = -120;
Δ = b2-4ac
Δ = 72-4·1·(-120)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*1}=\frac{-30}{2}
=-15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*1}=\frac{16}{2}
=8
$