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(2x+x-5)(10x+5)=(5x+5)(3x-1)
We move all terms to the left:
(2x+x-5)(10x+5)-((5x+5)(3x-1))=0
We add all the numbers together, and all the variables
(3x-5)(10x+5)-((5x+5)(3x-1))=0
We multiply parentheses ..
(+30x^2+15x-50x-25)-((5x+5)(3x-1))=0
We calculate terms in parentheses: -((5x+5)(3x-1)), so:We get rid of parentheses
(5x+5)(3x-1)
We multiply parentheses ..
(+15x^2-5x+15x-5)
We get rid of parentheses
15x^2-5x+15x-5
We add all the numbers together, and all the variables
15x^2+10x-5
Back to the equation:
-(15x^2+10x-5)
30x^2-15x^2+15x-50x-10x-25+5=0
We add all the numbers together, and all the variables
15x^2-45x-20=0
a = 15; b = -45; c = -20;
Δ = b2-4ac
Δ = -452-4·15·(-20)
Δ = 3225
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3225}=\sqrt{25*129}=\sqrt{25}*\sqrt{129}=5\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-5\sqrt{129}}{2*15}=\frac{45-5\sqrt{129}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+5\sqrt{129}}{2*15}=\frac{45+5\sqrt{129}}{30} $
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