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.6x^2+10x+3=0
a = .6; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·.6·3
Δ = 92.8
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{-(10)-\sqrt{92.8}}{2*.6}=\frac{-10-\sqrt{92.8}}{1.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{92.8}}{2*.6}=\frac{-10+\sqrt{92.8}}{1.2} $
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