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18x^2+98x-1296=0
a = 18; b = 98; c = -1296;
Δ = b2-4ac
Δ = 982-4·18·(-1296)
Δ = 102916
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{102916}=\sqrt{4*25729}=\sqrt{4}*\sqrt{25729}=2\sqrt{25729}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-2\sqrt{25729}}{2*18}=\frac{-98-2\sqrt{25729}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+2\sqrt{25729}}{2*18}=\frac{-98+2\sqrt{25729}}{36} $
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