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(2x-1)(x+1)-(x-3)(x+5)=20
We move all terms to the left:
(2x-1)(x+1)-(x-3)(x+5)-(20)=0
We multiply parentheses ..
(+2x^2+2x-1x-1)-(x-3)(x+5)-20=0
We get rid of parentheses
2x^2+2x-1x-(x-3)(x+5)-1-20=0
We multiply parentheses ..
2x^2-(+x^2+5x-3x-15)+2x-1x-1-20=0
We add all the numbers together, and all the variables
2x^2-(+x^2+5x-3x-15)+x-21=0
We get rid of parentheses
2x^2-x^2-5x+3x+x+15-21=0
We add all the numbers together, and all the variables
x^2-1x-6=0
a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2} =3 $
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