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(2x-4)(3x+9)=6
We move all terms to the left:
(2x-4)(3x+9)-(6)=0
We multiply parentheses ..
(+6x^2+18x-12x-36)-6=0
We get rid of parentheses
6x^2+18x-12x-36-6=0
We add all the numbers together, and all the variables
6x^2+6x-42=0
a = 6; b = 6; c = -42;
Δ = b2-4ac
Δ = 62-4·6·(-42)
Δ = 1044
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{29}}{2*6}=\frac{-6-6\sqrt{29}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{29}}{2*6}=\frac{-6+6\sqrt{29}}{12} $
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