5(21)=(x+3)(3x+6)

Solution for 5(21)=(x+3)(3x+6) equation:

5(21)=(x+3)(3x+6)

We move all terms to the left:

5(21)-((x+3)(3x+6))=0

We multiply parentheses ..

-((+3x^2+6x+9x+18))+521=0


We calculate terms in parentheses: -((+3x^2+6x+9x+18)), so:

(+3x^2+6x+9x+18)

We get rid of parentheses

3x^2+6x+9x+18

We add all the numbers together, and all the variables

3x^2+15x+18

Back to the equation:

-(3x^2+15x+18)


We get rid of parentheses

-3x^2-15x-18+521=0

We add all the numbers together, and all the variables

-3x^2-15x+503=0

a = -3; b = -15; c = +503;
Δ = b2-4ac
Δ = -152-4·(-3)·503
Δ = 6261
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{-(-15)-\sqrt{6261}}{2*-3}=\frac{15-\sqrt{6261}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{6261}}{2*-3}=\frac{15+\sqrt{6261}}{-6}
$