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(2x+3)(3x+1)=11+30
We move all terms to the left:
(2x+3)(3x+1)-(11+30)=0
We add all the numbers together, and all the variables
(2x+3)(3x+1)-41=0
We multiply parentheses ..
(+6x^2+2x+9x+3)-41=0
We get rid of parentheses
6x^2+2x+9x+3-41=0
We add all the numbers together, and all the variables
6x^2+11x-38=0
a = 6; b = 11; c = -38;
Δ = b2-4ac
Δ = 112-4·6·(-38)
Δ = 1033
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{-(11)-\sqrt{1033}}{2*6}=\frac{-11-\sqrt{1033}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1033}}{2*6}=\frac{-11+\sqrt{1033}}{12} $
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