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3(2x+4)-2x(x+1)=3(2-4x)
We move all terms to the left:
3(2x+4)-2x(x+1)-(3(2-4x))=0
We add all the numbers together, and all the variables
3(2x+4)-2x(x+1)-(3(-4x+2))=0
We multiply parentheses
-2x^2+6x-2x-(3(-4x+2))+12=0
We calculate terms in parentheses: -(3(-4x+2)), so:We add all the numbers together, and all the variables
3(-4x+2)
We multiply parentheses
-12x+6
Back to the equation:
-(-12x+6)
-2x^2+4x-(-12x+6)+12=0
We get rid of parentheses
-2x^2+4x+12x-6+12=0
We add all the numbers together, and all the variables
-2x^2+16x+6=0
a = -2; b = 16; c = +6;
Δ = b2-4ac
Δ = 162-4·(-2)·6
Δ = 304
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{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{19}}{2*-2}=\frac{-16-4\sqrt{19}}{-4}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{19}}{2*-2}=\frac{-16+4\sqrt{19}}{-4}
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