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2a^2-3a=3
We move all terms to the left:
2a^2-3a-(3)=0
a = 2; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·2·(-3)
Δ = 33
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*2}=\frac{3-\sqrt{33}}{4} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*2}=\frac{3+\sqrt{33}}{4} $
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