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(3x+1)(2x-7)=19
We move all terms to the left:
(3x+1)(2x-7)-(19)=0
We multiply parentheses ..
(+6x^2-21x+2x-7)-19=0
We get rid of parentheses
6x^2-21x+2x-7-19=0
We add all the numbers together, and all the variables
6x^2-19x-26=0
a = 6; b = -19; c = -26;
Δ = b2-4ac
Δ = -192-4·6·(-26)
Δ = 985
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{-(-19)-\sqrt{985}}{2*6}=\frac{19-\sqrt{985}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{985}}{2*6}=\frac{19+\sqrt{985}}{12} $
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