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3.5x^2+12x+6=0
a = 3.5; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·3.5·6
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{15}}{2*3.5}=\frac{-12-2\sqrt{15}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{15}}{2*3.5}=\frac{-12+2\sqrt{15}}{7} $
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