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x^2+150x-500=0
a = 1; b = 150; c = -500;
Δ = b2-4ac
Δ = 1502-4·1·(-500)
Δ = 24500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24500}=\sqrt{4900*5}=\sqrt{4900}*\sqrt{5}=70\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-70\sqrt{5}}{2*1}=\frac{-150-70\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+70\sqrt{5}}{2*1}=\frac{-150+70\sqrt{5}}{2} $
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