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2x^2+2x=8-6x
We move all terms to the left:
2x^2+2x-(8-6x)=0
We add all the numbers together, and all the variables
2x^2+2x-(-6x+8)=0
We get rid of parentheses
2x^2+2x+6x-8=0
We add all the numbers together, and all the variables
2x^2+8x-8=0
a = 2; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·2·(-8)
Δ = 128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*2}=\frac{-8-8\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*2}=\frac{-8+8\sqrt{2}}{4} $
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