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(2x-4)(-3x-5)=0
We multiply parentheses ..
(-6x^2-10x+12x+20)=0
We get rid of parentheses
-6x^2-10x+12x+20=0
We add all the numbers together, and all the variables
-6x^2+2x+20=0
a = -6; b = 2; c = +20;
Δ = b2-4ac
Δ = 22-4·(-6)·20
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*-6}=\frac{-24}{-12} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*-6}=\frac{20}{-12} =-1+2/3 $
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