9x^2-72x+98=0

Solution for 9x^2-72x+98=0 equation:

9x^2-72x+98=0

a = 9; b = -72; c = +98;
Δ = b2-4ac
Δ = -722-4·9·98
Δ = 1656
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:

$\sqrt{\Delta}=\sqrt{1656}=\sqrt{36*46}=\sqrt{36}*\sqrt{46}=6\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-6\sqrt{46}}{2*9}=\frac{72-6\sqrt{46}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+6\sqrt{46}}{2*9}=\frac{72+6\sqrt{46}}{18}
$