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10x^2+12x-6=0
a = 10; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·10·(-6)
Δ = 384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{6}}{2*10}=\frac{-12-8\sqrt{6}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{6}}{2*10}=\frac{-12+8\sqrt{6}}{20} $
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