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(2x+3)(2x+4)=0
We multiply parentheses ..
(+4x^2+8x+6x+12)=0
We get rid of parentheses
4x^2+8x+6x+12=0
We add all the numbers together, and all the variables
4x^2+14x+12=0
a = 4; b = 14; c = +12;
Δ = b2-4ac
Δ = 142-4·4·12
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*4}=\frac{-12}{8} =-1+1/2 $
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