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15x^2-18x-68=0
a = 15; b = -18; c = -68;
Δ = b2-4ac
Δ = -182-4·15·(-68)
Δ = 4404
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4404}=\sqrt{4*1101}=\sqrt{4}*\sqrt{1101}=2\sqrt{1101}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{1101}}{2*15}=\frac{18-2\sqrt{1101}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{1101}}{2*15}=\frac{18+2\sqrt{1101}}{30} $
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