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3r^2+3=195
We move all terms to the left:
3r^2+3-(195)=0
We add all the numbers together, and all the variables
3r^2-192=0
a = 3; b = 0; c = -192;
Δ = b2-4ac
Δ = 02-4·3·(-192)
Δ = 2304
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{2304}=48$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*3}=\frac{-48}{6} =-8 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*3}=\frac{48}{6} =8 $
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