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(3x-5)(2x-10)=0
We multiply parentheses ..
(+6x^2-30x-10x+50)=0
We get rid of parentheses
6x^2-30x-10x+50=0
We add all the numbers together, and all the variables
6x^2-40x+50=0
a = 6; b = -40; c = +50;
Δ = b2-4ac
Δ = -402-4·6·50
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*6}=\frac{20}{12} =1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*6}=\frac{60}{12} =5 $
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