9x^2+42x+45=(3x-5)(2x+1)

Solution for 9x^2+42x+45=(3x-5)(2x+1) equation:

9x^2+42x+45=(3x-5)(2x+1)

We move all terms to the left:

9x^2+42x+45-((3x-5)(2x+1))=0

We multiply parentheses ..

9x^2-((+6x^2+3x-10x-5))+42x+45=0


We calculate terms in parentheses: -((+6x^2+3x-10x-5)), so:

(+6x^2+3x-10x-5)

We get rid of parentheses

6x^2+3x-10x-5

We add all the numbers together, and all the variables

6x^2-7x-5

Back to the equation:

-(6x^2-7x-5)


We add all the numbers together, and all the variables

9x^2+42x-(6x^2-7x-5)+45=0

We get rid of parentheses

9x^2-6x^2+42x+7x+5+45=0

We add all the numbers together, and all the variables

3x^2+49x+50=0

a = 3; b = 49; c = +50;
Δ = b2-4ac
Δ = 492-4·3·50
Δ = 1801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-\sqrt{1801}}{2*3}=\frac{-49-\sqrt{1801}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+\sqrt{1801}}{2*3}=\frac{-49+\sqrt{1801}}{6}
$