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x^2-248x-1500=0
a = 1; b = -248; c = -1500;
Δ = b2-4ac
Δ = -2482-4·1·(-1500)
Δ = 67504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{67504}=\sqrt{16*4219}=\sqrt{16}*\sqrt{4219}=4\sqrt{4219}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-248)-4\sqrt{4219}}{2*1}=\frac{248-4\sqrt{4219}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-248)+4\sqrt{4219}}{2*1}=\frac{248+4\sqrt{4219}}{2} $
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