(7x-99)(3x-31)=180

Solution for (7x-99)(3x-31)=180 equation:

(7x-99)(3x-31)=180

We move all terms to the left:

(7x-99)(3x-31)-(180)=0

We multiply parentheses ..

(+21x^2-217x-297x+3069)-180=0

We get rid of parentheses

21x^2-217x-297x+3069-180=0

We add all the numbers together, and all the variables

21x^2-514x+2889=0

a = 21; b = -514; c = +2889;
Δ = b2-4ac
Δ = -5142-4·21·2889
Δ = 21520
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:

$\sqrt{\Delta}=\sqrt{21520}=\sqrt{16*1345}=\sqrt{16}*\sqrt{1345}=4\sqrt{1345}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-514)-4\sqrt{1345}}{2*21}=\frac{514-4\sqrt{1345}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-514)+4\sqrt{1345}}{2*21}=\frac{514+4\sqrt{1345}}{42}
$