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x^2+10+14x=0
a = 1; b = 14; c = +10;
Δ = b2-4ac
Δ = 142-4·1·10
Δ = 156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{39}}{2*1}=\frac{-14-2\sqrt{39}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{39}}{2*1}=\frac{-14+2\sqrt{39}}{2} $
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