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3x^2+3x-1900=0
a = 3; b = 3; c = -1900;
Δ = b2-4ac
Δ = 32-4·3·(-1900)
Δ = 22809
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{22809}}{2*3}=\frac{-3-\sqrt{22809}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{22809}}{2*3}=\frac{-3+\sqrt{22809}}{6} $
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