(3x+12)(3x-18)=5

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Solution for (3x+12)(3x-18)=5 equation:



(3x+12)(3x-18)=5
We move all terms to the left:
(3x+12)(3x-18)-(5)=0
We multiply parentheses ..
(+9x^2-54x+36x-216)-5=0
We get rid of parentheses
9x^2-54x+36x-216-5=0
We add all the numbers together, and all the variables
9x^2-18x-221=0
a = 9; b = -18; c = -221;
Δ = b2-4ac
Δ = -182-4·9·(-221)
Δ = 8280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{8280}=\sqrt{36*230}=\sqrt{36}*\sqrt{230}=6\sqrt{230}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{230}}{2*9}=\frac{18-6\sqrt{230}}{18} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{230}}{2*9}=\frac{18+6\sqrt{230}}{18} $

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