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(2x+5)(3x+5)=6
We move all terms to the left:
(2x+5)(3x+5)-(6)=0
We multiply parentheses ..
(+6x^2+10x+15x+25)-6=0
We get rid of parentheses
6x^2+10x+15x+25-6=0
We add all the numbers together, and all the variables
6x^2+25x+19=0
a = 6; b = 25; c = +19;
Δ = b2-4ac
Δ = 252-4·6·19
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-13}{2*6}=\frac{-38}{12} =-3+1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+13}{2*6}=\frac{-12}{12} =-1 $
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