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(6x-4)(4x+8)=0
We multiply parentheses ..
(+24x^2+48x-16x-32)=0
We get rid of parentheses
24x^2+48x-16x-32=0
We add all the numbers together, and all the variables
24x^2+32x-32=0
a = 24; b = 32; c = -32;
Δ = b2-4ac
Δ = 322-4·24·(-32)
Δ = 4096
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{4096}=64$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-64}{2*24}=\frac{-96}{48} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+64}{2*24}=\frac{32}{48} =2/3 $
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