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6=r2+4
We move all terms to the left:
6-(r2+4)=0
We add all the numbers together, and all the variables
-(+r^2+4)+6=0
We get rid of parentheses
-r^2-4+6=0
We add all the numbers together, and all the variables
-1r^2+2=0
a = -1; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-1)·2
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-1}=\frac{0-2\sqrt{2}}{-2} =-\frac{2\sqrt{2}}{-2} =-\frac{\sqrt{2}}{-1} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-1}=\frac{0+2\sqrt{2}}{-2} =\frac{2\sqrt{2}}{-2} =\frac{\sqrt{2}}{-1} $
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