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10x^2-50x-40=0
a = 10; b = -50; c = -40;
Δ = b2-4ac
Δ = -502-4·10·(-40)
Δ = 4100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4100}=\sqrt{100*41}=\sqrt{100}*\sqrt{41}=10\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{41}}{2*10}=\frac{50-10\sqrt{41}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{41}}{2*10}=\frac{50+10\sqrt{41}}{20} $
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