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3a^2-5+2=0
We add all the numbers together, and all the variables
3a^2-3=0
a = 3; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·3·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*3}=\frac{-6}{6} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*3}=\frac{6}{6} =1 $
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