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6x^2-10x-50=0
a = 6; b = -10; c = -50;
Δ = b2-4ac
Δ = -102-4·6·(-50)
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{13}}{2*6}=\frac{10-10\sqrt{13}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{13}}{2*6}=\frac{10+10\sqrt{13}}{12} $
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