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(3x+5)(4x+8)=0
We multiply parentheses ..
(+12x^2+24x+20x+40)=0
We get rid of parentheses
12x^2+24x+20x+40=0
We add all the numbers together, and all the variables
12x^2+44x+40=0
a = 12; b = 44; c = +40;
Δ = b2-4ac
Δ = 442-4·12·40
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4}{2*12}=\frac{-48}{24} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4}{2*12}=\frac{-40}{24} =-1+2/3 $
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