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(6x-4)(3x+4)=0
We multiply parentheses ..
(+18x^2+24x-12x-16)=0
We get rid of parentheses
18x^2+24x-12x-16=0
We add all the numbers together, and all the variables
18x^2+12x-16=0
a = 18; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·18·(-16)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*18}=\frac{-48}{36} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*18}=\frac{24}{36} =2/3 $
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