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(5x+3)(4x+3)=0
We multiply parentheses ..
(+20x^2+15x+12x+9)=0
We get rid of parentheses
20x^2+15x+12x+9=0
We add all the numbers together, and all the variables
20x^2+27x+9=0
a = 20; b = 27; c = +9;
Δ = b2-4ac
Δ = 272-4·20·9
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3}{2*20}=\frac{-30}{40} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3}{2*20}=\frac{-24}{40} =-3/5 $
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