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7x^2+42x+9=0
a = 7; b = 42; c = +9;
Δ = b2-4ac
Δ = 422-4·7·9
Δ = 1512
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1512}=\sqrt{36*42}=\sqrt{36}*\sqrt{42}=6\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{42}}{2*7}=\frac{-42-6\sqrt{42}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{42}}{2*7}=\frac{-42+6\sqrt{42}}{14} $
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