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6x^2+12x=2x+8
We move all terms to the left:
6x^2+12x-(2x+8)=0
We get rid of parentheses
6x^2+12x-2x-8=0
We add all the numbers together, and all the variables
6x^2+10x-8=0
a = 6; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·6·(-8)
Δ = 292
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{73}}{2*6}=\frac{-10-2\sqrt{73}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{73}}{2*6}=\frac{-10+2\sqrt{73}}{12} $
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