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(3x^2)-(350x)+7500=0
a = 3; b = -350; c = +7500;
Δ = b2-4ac
Δ = -3502-4·3·7500
Δ = 32500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32500}=\sqrt{2500*13}=\sqrt{2500}*\sqrt{13}=50\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-350)-50\sqrt{13}}{2*3}=\frac{350-50\sqrt{13}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-350)+50\sqrt{13}}{2*3}=\frac{350+50\sqrt{13}}{6} $
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