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10x^2+28x+7=0
a = 10; b = 28; c = +7;
Δ = b2-4ac
Δ = 282-4·10·7
Δ = 504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-6\sqrt{14}}{2*10}=\frac{-28-6\sqrt{14}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+6\sqrt{14}}{2*10}=\frac{-28+6\sqrt{14}}{20} $
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