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-6-2r^2=-26
We move all terms to the left:
-6-2r^2-(-26)=0
We add all the numbers together, and all the variables
-2r^2+20=0
a = -2; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-2)·20
Δ = 160
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-2}=\frac{0-4\sqrt{10}}{-4} =-\frac{4\sqrt{10}}{-4} =-\frac{\sqrt{10}}{-1} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-2}=\frac{0+4\sqrt{10}}{-4} =\frac{4\sqrt{10}}{-4} =\frac{\sqrt{10}}{-1} $
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