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18x^2+20x-13=0
a = 18; b = 20; c = -13;
Δ = b2-4ac
Δ = 202-4·18·(-13)
Δ = 1336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1336}=\sqrt{4*334}=\sqrt{4}*\sqrt{334}=2\sqrt{334}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{334}}{2*18}=\frac{-20-2\sqrt{334}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{334}}{2*18}=\frac{-20+2\sqrt{334}}{36} $
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