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2x^2-10x-199=0
a = 2; b = -10; c = -199;
Δ = b2-4ac
Δ = -102-4·2·(-199)
Δ = 1692
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1692}=\sqrt{36*47}=\sqrt{36}*\sqrt{47}=6\sqrt{47}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{47}}{2*2}=\frac{10-6\sqrt{47}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{47}}{2*2}=\frac{10+6\sqrt{47}}{4} $
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