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7x^2+28x+16=0
a = 7; b = 28; c = +16;
Δ = b2-4ac
Δ = 282-4·7·16
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{21}}{2*7}=\frac{-28-4\sqrt{21}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{21}}{2*7}=\frac{-28+4\sqrt{21}}{14} $
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