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3x^2+15x-250=0
a = 3; b = 15; c = -250;
Δ = b2-4ac
Δ = 152-4·3·(-250)
Δ = 3225
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3225}=\sqrt{25*129}=\sqrt{25}*\sqrt{129}=5\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{129}}{2*3}=\frac{-15-5\sqrt{129}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{129}}{2*3}=\frac{-15+5\sqrt{129}}{6} $
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