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10x^2+50x+14.000=0
We add all the numbers together, and all the variables
10x^2+50x+14=0
a = 10; b = 50; c = +14;
Δ = b2-4ac
Δ = 502-4·10·14
Δ = 1940
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1940}=\sqrt{4*485}=\sqrt{4}*\sqrt{485}=2\sqrt{485}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{485}}{2*10}=\frac{-50-2\sqrt{485}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{485}}{2*10}=\frac{-50+2\sqrt{485}}{20} $
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