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4r^2-62=2
We move all terms to the left:
4r^2-62-(2)=0
We add all the numbers together, and all the variables
4r^2-64=0
a = 4; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·4·(-64)
Δ = 1024
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{1024}=32$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*4}=\frac{-32}{8} =-4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*4}=\frac{32}{8} =4 $
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