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(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^2-7x=0
a = 6; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·6·0
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*6}=\frac{14}{12} =1+1/6 $
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