(2x+1)(3x+2)=100

Solution for (2x+1)(3x+2)=100 equation:

(2x+1)(3x+2)=100

We move all terms to the left:

(2x+1)(3x+2)-(100)=0

We multiply parentheses ..

(+6x^2+4x+3x+2)-100=0

We get rid of parentheses

6x^2+4x+3x+2-100=0

We add all the numbers together, and all the variables

6x^2+7x-98=0

a = 6; b = 7; c = -98;
Δ = b2-4ac
Δ = 72-4·6·(-98)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-49}{2*6}=\frac{-56}{12}
=-4+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+49}{2*6}=\frac{42}{12}
=3+1/2
$