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(2x+3)(3x+8)=0
We multiply parentheses ..
(+6x^2+16x+9x+24)=0
We get rid of parentheses
6x^2+16x+9x+24=0
We add all the numbers together, and all the variables
6x^2+25x+24=0
a = 6; b = 25; c = +24;
Δ = b2-4ac
Δ = 252-4·6·24
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-7}{2*6}=\frac{-32}{12} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+7}{2*6}=\frac{-18}{12} =-1+1/2 $
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