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3x^2+19x-8.9=0
a = 3; b = 19; c = -8.9;
Δ = b2-4ac
Δ = 192-4·3·(-8.9)
Δ = 467.8
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{-(19)-\sqrt{467.8}}{2*3}=\frac{-19-\sqrt{467.8}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{467.8}}{2*3}=\frac{-19+\sqrt{467.8}}{6} $
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