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(2x+5)(3x+4)+1=9
We move all terms to the left:
(2x+5)(3x+4)+1-(9)=0
We add all the numbers together, and all the variables
(2x+5)(3x+4)-8=0
We multiply parentheses ..
(+6x^2+8x+15x+20)-8=0
We get rid of parentheses
6x^2+8x+15x+20-8=0
We add all the numbers together, and all the variables
6x^2+23x+12=0
a = 6; b = 23; c = +12;
Δ = b2-4ac
Δ = 232-4·6·12
Δ = 241
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{-(23)-\sqrt{241}}{2*6}=\frac{-23-\sqrt{241}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{241}}{2*6}=\frac{-23+\sqrt{241}}{12} $
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