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3x^2+2x-800=0
a = 3; b = 2; c = -800;
Δ = b2-4ac
Δ = 22-4·3·(-800)
Δ = 9604
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9604}=98$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-98}{2*3}=\frac{-100}{6} =-16+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+98}{2*3}=\frac{96}{6} =16 $
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