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3x^2+5x-16.5=0
a = 3; b = 5; c = -16.5;
Δ = b2-4ac
Δ = 52-4·3·(-16.5)
Δ = 223
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{223}}{2*3}=\frac{-5-\sqrt{223}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{223}}{2*3}=\frac{-5+\sqrt{223}}{6} $
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