9x^2+96x+112=0

Solution for 9x^2+96x+112=0 equation:

9x^2+96x+112=0

a = 9; b = 96; c = +112;
Δ = b2-4ac
Δ = 962-4·9·112
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-72}{2*9}=\frac{-168}{18}
=-9+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+72}{2*9}=\frac{-24}{18}
=-1+1/3
$