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(3x-1)(2x+3)=39
We move all terms to the left:
(3x-1)(2x+3)-(39)=0
We multiply parentheses ..
(+6x^2+9x-2x-3)-39=0
We get rid of parentheses
6x^2+9x-2x-3-39=0
We add all the numbers together, and all the variables
6x^2+7x-42=0
a = 6; b = 7; c = -42;
Δ = b2-4ac
Δ = 72-4·6·(-42)
Δ = 1057
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{-(7)-\sqrt{1057}}{2*6}=\frac{-7-\sqrt{1057}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{1057}}{2*6}=\frac{-7+\sqrt{1057}}{12} $
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