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(3x+5)(3x+4)=0
We multiply parentheses ..
(+9x^2+12x+15x+20)=0
We get rid of parentheses
9x^2+12x+15x+20=0
We add all the numbers together, and all the variables
9x^2+27x+20=0
a = 9; b = 27; c = +20;
Δ = b2-4ac
Δ = 272-4·9·20
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3}{2*9}=\frac{-30}{18} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3}{2*9}=\frac{-24}{18} =-1+1/3 $
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