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4(x+3)=6(x-1)x=
We move all terms to the left:
4(x+3)-(6(x-1)x)=0
We multiply parentheses
4x-(6(x-1)x)+12=0
We calculate terms in parentheses: -(6(x-1)x), so:We get rid of parentheses
6(x-1)x
We multiply parentheses
6x^2-6x
Back to the equation:
-(6x^2-6x)
-6x^2+4x+6x+12=0
We add all the numbers together, and all the variables
-6x^2+10x+12=0
a = -6; b = 10; c = +12;
Δ = b2-4ac
Δ = 102-4·(-6)·12
Δ = 388
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{97}}{2*-6}=\frac{-10-2\sqrt{97}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{97}}{2*-6}=\frac{-10+2\sqrt{97}}{-12} $
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