9x^2+149x+590=0

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Solution for 9x^2+149x+590=0 equation:



9x^2+149x+590=0
a = 9; b = 149; c = +590;
Δ = b2-4ac
Δ = 1492-4·9·590
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(149)-31}{2*9}=\frac{-180}{18} =-10 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(149)+31}{2*9}=\frac{-118}{18} =-6+5/9 $

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