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(2x+6x^2)=(4x+3)+2
We move all terms to the left:
(2x+6x^2)-((4x+3)+2)=0
We get rid of parentheses
6x^2+2x-((4x+3)+2)=0
We calculate terms in parentheses: -((4x+3)+2), so:We get rid of parentheses
(4x+3)+2
We get rid of parentheses
4x+3+2
We add all the numbers together, and all the variables
4x+5
Back to the equation:
-(4x+5)
6x^2+2x-4x-5=0
We add all the numbers together, and all the variables
6x^2-2x-5=0
a = 6; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·6·(-5)
Δ = 124
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{31}}{2*6}=\frac{2-2\sqrt{31}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{31}}{2*6}=\frac{2+2\sqrt{31}}{12} $
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