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(3x+5)(2x+3)=0
We multiply parentheses ..
(+6x^2+9x+10x+15)=0
We get rid of parentheses
6x^2+9x+10x+15=0
We add all the numbers together, and all the variables
6x^2+19x+15=0
a = 6; b = 19; c = +15;
Δ = b2-4ac
Δ = 192-4·6·15
Δ = 1
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*6}=\frac{-20}{12} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*6}=\frac{-18}{12} =-1+1/2 $
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