(3x+1)(x-7)=100

Solution for (3x+1)(x-7)=100 equation:

(3x+1)(x-7)=100

We move all terms to the left:

(3x+1)(x-7)-(100)=0

We multiply parentheses ..

(+3x^2-21x+x-7)-100=0

We get rid of parentheses

3x^2-21x+x-7-100=0

We add all the numbers together, and all the variables

3x^2-20x-107=0

a = 3; b = -20; c = -107;
Δ = b2-4ac
Δ = -202-4·3·(-107)
Δ = 1684
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:

$\sqrt{\Delta}=\sqrt{1684}=\sqrt{4*421}=\sqrt{4}*\sqrt{421}=2\sqrt{421}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{421}}{2*3}=\frac{20-2\sqrt{421}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{421}}{2*3}=\frac{20+2\sqrt{421}}{6}
$