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(2x-3)(x+2)-(4x-2)(x-5)=-16

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Solution for (2x-3)(x+2)-(4x-2)(x-5)=-16 equation:



(2x-3)(x+2)-(4x-2)(x-5)=-16
We move all terms to the left:
(2x-3)(x+2)-(4x-2)(x-5)-(-16)=0
We add all the numbers together, and all the variables
(2x-3)(x+2)-(4x-2)(x-5)+16=0
We multiply parentheses ..
(+2x^2+4x-3x-6)-(4x-2)(x-5)+16=0
We get rid of parentheses
2x^2+4x-3x-(4x-2)(x-5)-6+16=0
We multiply parentheses ..
2x^2-(+4x^2-20x-2x+10)+4x-3x-6+16=0
We add all the numbers together, and all the variables
2x^2-(+4x^2-20x-2x+10)+x+10=0
We get rid of parentheses
2x^2-4x^2+20x+2x+x-10+10=0
We add all the numbers together, and all the variables
-2x^2+23x=0
a = -2; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·(-2)·0
Δ = 529
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{529}=23
x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*-2}=\frac{-46}{-4} =11+1/2
x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*-2}=\frac{0}{-4} =0

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