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(2x+1)(3x+5)=5
We move all terms to the left:
(2x+1)(3x+5)-(5)=0
We multiply parentheses ..
(+6x^2+10x+3x+5)-5=0
We get rid of parentheses
6x^2+10x+3x+5-5=0
We add all the numbers together, and all the variables
6x^2+13x=0
a = 6; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·6·0
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*6}=\frac{-26}{12} =-2+1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*6}=\frac{0}{12} =0 $
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