(2x+3)(1x+2)=45

Solution for (2x+3)(1x+2)=45 equation:

(2x+3)(1x+2)=45

We move all terms to the left:

(2x+3)(1x+2)-(45)=0

We add all the numbers together, and all the variables

(2x+3)(x+2)-45=0

We multiply parentheses ..

(+2x^2+4x+3x+6)-45=0

We get rid of parentheses

2x^2+4x+3x+6-45=0

We add all the numbers together, and all the variables

2x^2+7x-39=0

a = 2; b = 7; c = -39;
Δ = b2-4ac
Δ = 72-4·2·(-39)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-19}{2*2}=\frac{-26}{4}
=-6+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+19}{2*2}=\frac{12}{4}
=3
$