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3x^2-16x+2.5=0
a = 3; b = -16; c = +2.5;
Δ = b2-4ac
Δ = -162-4·3·2.5
Δ = 226
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{-(-16)-\sqrt{226}}{2*3}=\frac{16-\sqrt{226}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{226}}{2*3}=\frac{16+\sqrt{226}}{6} $
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