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(2x+4)(3x+4)=0
We multiply parentheses ..
(+6x^2+8x+12x+16)=0
We get rid of parentheses
6x^2+8x+12x+16=0
We add all the numbers together, and all the variables
6x^2+20x+16=0
a = 6; b = 20; c = +16;
Δ = b2-4ac
Δ = 202-4·6·16
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*6}=\frac{-24}{12} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*6}=\frac{-16}{12} =-1+1/3 $
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