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(2x-5)(3x=10)
We move all terms to the left:
(2x-5)(3x-(10))=0
We multiply parentheses ..
(+6x^2-20x-15x+50)=0
We get rid of parentheses
6x^2-20x-15x+50=0
We add all the numbers together, and all the variables
6x^2-35x+50=0
a = 6; b = -35; c = +50;
Δ = b2-4ac
Δ = -352-4·6·50
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5}{2*6}=\frac{30}{12} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5}{2*6}=\frac{40}{12} =3+1/3 $
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