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5x^2+x-403=0
a = 5; b = 1; c = -403;
Δ = b2-4ac
Δ = 12-4·5·(-403)
Δ = 8061
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{-(1)-\sqrt{8061}}{2*5}=\frac{-1-\sqrt{8061}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{8061}}{2*5}=\frac{-1+\sqrt{8061}}{10} $
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