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3x^2+19x-390=0
a = 3; b = 19; c = -390;
Δ = b2-4ac
Δ = 192-4·3·(-390)
Δ = 5041
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{5041}=71$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-71}{2*3}=\frac{-90}{6} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+71}{2*3}=\frac{52}{6} =8+2/3 $
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