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3x-x+4=x(-2+4)
We move all terms to the left:
3x-x+4-(x(-2+4))=0
We add all the numbers together, and all the variables
3x-x-(x2)+4=0
We add all the numbers together, and all the variables
-1x^2+2x+4=0
a = -1; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-1)·4
Δ = 20
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*-1}=\frac{-2-2\sqrt{5}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*-1}=\frac{-2+2\sqrt{5}}{-2} $
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