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(2x+2)(3x-8)=0
We multiply parentheses ..
(+6x^2-16x+6x-16)=0
We get rid of parentheses
6x^2-16x+6x-16=0
We add all the numbers together, and all the variables
6x^2-10x-16=0
a = 6; b = -10; c = -16;
Δ = b2-4ac
Δ = -102-4·6·(-16)
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*6}=\frac{-12}{12} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*6}=\frac{32}{12} =2+2/3 $
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