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-3r^2-r-6=-4r^2
We move all terms to the left:
-3r^2-r-6-(-4r^2)=0
We add all the numbers together, and all the variables
-3r^2-(-4r^2)-1r-6=0
We get rid of parentheses
-3r^2+4r^2-1r-6=0
We add all the numbers together, and all the variables
r^2-1r-6=0
a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2} =3 $
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