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

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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} $

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