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2x^2-9x-195=0
a = 2; b = -9; c = -195;
Δ = b2-4ac
Δ = -92-4·2·(-195)
Δ = 1641
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{-(-9)-\sqrt{1641}}{2*2}=\frac{9-\sqrt{1641}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{1641}}{2*2}=\frac{9+\sqrt{1641}}{4} $
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