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x^2+60x-3000=0
a = 1; b = 60; c = -3000;
Δ = b2-4ac
Δ = 602-4·1·(-3000)
Δ = 15600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15600}=\sqrt{400*39}=\sqrt{400}*\sqrt{39}=20\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{39}}{2*1}=\frac{-60-20\sqrt{39}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{39}}{2*1}=\frac{-60+20\sqrt{39}}{2} $
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