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10x^2+40x+7=0
a = 10; b = 40; c = +7;
Δ = b2-4ac
Δ = 402-4·10·7
Δ = 1320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1320}=\sqrt{4*330}=\sqrt{4}*\sqrt{330}=2\sqrt{330}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{330}}{2*10}=\frac{-40-2\sqrt{330}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{330}}{2*10}=\frac{-40+2\sqrt{330}}{20} $
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