9x^2+198x+1062=0

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Solution for 9x^2+198x+1062=0 equation:



9x^2+198x+1062=0
a = 9; b = 198; c = +1062;
Δ = b2-4ac
Δ = 1982-4·9·1062
Δ = 972
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:
$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(198)-18\sqrt{3}}{2*9}=\frac{-198-18\sqrt{3}}{18} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(198)+18\sqrt{3}}{2*9}=\frac{-198+18\sqrt{3}}{18} $

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