9x^2+41x+46=0

Solution for 9x^2+41x+46=0 equation:

9x^2+41x+46=0

a = 9; b = 41; c = +46;
Δ = b2-4ac
Δ = 412-4·9·46
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-5}{2*9}=\frac{-46}{18}
=-2+5/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+5}{2*9}=\frac{-36}{18}
=-2
$