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7a^2+28a+24=0
a = 7; b = 28; c = +24;
Δ = b2-4ac
Δ = 282-4·7·24
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{7}}{2*7}=\frac{-28-4\sqrt{7}}{14} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{7}}{2*7}=\frac{-28+4\sqrt{7}}{14} $
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