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18x^2+12x-11=0
a = 18; b = 12; c = -11;
Δ = b2-4ac
Δ = 122-4·18·(-11)
Δ = 936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{26}}{2*18}=\frac{-12-6\sqrt{26}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{26}}{2*18}=\frac{-12+6\sqrt{26}}{36} $
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