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15x^2+50x+6=0
a = 15; b = 50; c = +6;
Δ = b2-4ac
Δ = 502-4·15·6
Δ = 2140
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2140}=\sqrt{4*535}=\sqrt{4}*\sqrt{535}=2\sqrt{535}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{535}}{2*15}=\frac{-50-2\sqrt{535}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{535}}{2*15}=\frac{-50+2\sqrt{535}}{30} $
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