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(3x+1)(2x-6)=135
We move all terms to the left:
(3x+1)(2x-6)-(135)=0
We multiply parentheses ..
(+6x^2-18x+2x-6)-135=0
We get rid of parentheses
6x^2-18x+2x-6-135=0
We add all the numbers together, and all the variables
6x^2-16x-141=0
a = 6; b = -16; c = -141;
Δ = b2-4ac
Δ = -162-4·6·(-141)
Δ = 3640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3640}=\sqrt{4*910}=\sqrt{4}*\sqrt{910}=2\sqrt{910}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{910}}{2*6}=\frac{16-2\sqrt{910}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{910}}{2*6}=\frac{16+2\sqrt{910}}{12} $
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