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5x^2+3x-195=0
a = 5; b = 3; c = -195;
Δ = b2-4ac
Δ = 32-4·5·(-195)
Δ = 3909
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{-(3)-\sqrt{3909}}{2*5}=\frac{-3-\sqrt{3909}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{3909}}{2*5}=\frac{-3+\sqrt{3909}}{10} $
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