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3x^2+13x-430=0
a = 3; b = 13; c = -430;
Δ = b2-4ac
Δ = 132-4·3·(-430)
Δ = 5329
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{5329}=73$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-73}{2*3}=\frac{-86}{6} =-14+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+73}{2*3}=\frac{60}{6} =10 $
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