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x(2x-5)-3(2x-5)=0
We multiply parentheses
2x^2-5x-6x+15=0
We add all the numbers together, and all the variables
2x^2-11x+15=0
a = 2; b = -11; c = +15;
Δ = b2-4ac
Δ = -112-4·2·15
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*2}=\frac{10}{4} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*2}=\frac{12}{4} =3 $
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