9a^2-42a+26=0

Solution for 9a^2-42a+26=0 equation:

9a^2-42a+26=0

a = 9; b = -42; c = +26;
Δ = b2-4ac
Δ = -422-4·9·26
Δ = 828
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{828}=\sqrt{36*23}=\sqrt{36}*\sqrt{23}=6\sqrt{23}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{23}}{2*9}=\frac{42-6\sqrt{23}}{18}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{23}}{2*9}=\frac{42+6\sqrt{23}}{18}
$