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2x^2+4x=x2+10x-8
We move all terms to the left:
2x^2+4x-(x2+10x-8)=0
We add all the numbers together, and all the variables
2x^2-(+x^2+10x-8)+4x=0
We get rid of parentheses
2x^2-x^2-10x+4x+8=0
We add all the numbers together, and all the variables
x^2-6x+8=0
a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
Δ = 4
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2}{2*1}=\frac{4}{2} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*1}=\frac{8}{2} =4 $
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