202+x(3x+10)=540

Solution for 202+x(3x+10)=540 equation:

202+x(3x+10)=540

We move all terms to the left:

202+x(3x+10)-(540)=0

We add all the numbers together, and all the variables

x(3x+10)-338=0

We multiply parentheses

3x^2+10x-338=0

a = 3; b = 10; c = -338;
Δ = b2-4ac
Δ = 102-4·3·(-338)
Δ = 4156
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:

$\sqrt{\Delta}=\sqrt{4156}=\sqrt{4*1039}=\sqrt{4}*\sqrt{1039}=2\sqrt{1039}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1039}}{2*3}=\frac{-10-2\sqrt{1039}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1039}}{2*3}=\frac{-10+2\sqrt{1039}}{6}
$