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(2x+10)(x-2)=(x+10)(x+10)
We move all terms to the left:
(2x+10)(x-2)-((x+10)(x+10))=0
We multiply parentheses ..
(+2x^2-4x+10x-20)-((x+10)(x+10))=0
We calculate terms in parentheses: -((x+10)(x+10)), so:We get rid of parentheses
(x+10)(x+10)
We multiply parentheses ..
(+x^2+10x+10x+100)
We get rid of parentheses
x^2+10x+10x+100
We add all the numbers together, and all the variables
x^2+20x+100
Back to the equation:
-(x^2+20x+100)
2x^2-x^2-4x+10x-20x-20-100=0
We add all the numbers together, and all the variables
x^2-14x-120=0
a = 1; b = -14; c = -120;
Δ = b2-4ac
Δ = -142-4·1·(-120)
Δ = 676
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{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*1}=\frac{-12}{2} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*1}=\frac{40}{2} =20 $
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