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(2x+3)(9x-4)=3
We move all terms to the left:
(2x+3)(9x-4)-(3)=0
We multiply parentheses ..
(+18x^2-8x+27x-12)-3=0
We get rid of parentheses
18x^2-8x+27x-12-3=0
We add all the numbers together, and all the variables
18x^2+19x-15=0
a = 18; b = 19; c = -15;
Δ = b2-4ac
Δ = 192-4·18·(-15)
Δ = 1441
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{1441}}{2*18}=\frac{-19-\sqrt{1441}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1441}}{2*18}=\frac{-19+\sqrt{1441}}{36} $
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