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(3x+1)(2x-5)-(2x-5)(4x-3)=0
We multiply parentheses ..
(+6x^2-15x+2x-5)-(2x-5)(4x-3)=0
We get rid of parentheses
6x^2-15x+2x-(2x-5)(4x-3)-5=0
We multiply parentheses ..
6x^2-(+8x^2-6x-20x+15)-15x+2x-5=0
We add all the numbers together, and all the variables
6x^2-(+8x^2-6x-20x+15)-13x-5=0
We get rid of parentheses
6x^2-8x^2+6x+20x-13x-15-5=0
We add all the numbers together, and all the variables
-2x^2+13x-20=0
a = -2; b = 13; c = -20;
Δ = b2-4ac
Δ = 132-4·(-2)·(-20)
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*-2}=\frac{-16}{-4} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*-2}=\frac{-10}{-4} =2+1/2 $
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